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5.7: Linear Inequalities in One Variable

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• Page ID 49372

• Denny Burzynski & Wade Ellis, Jr.
• College of Southern Nevada via OpenStax CNX

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Inequalities

Relationships of Inequality

We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. Not all relationships need be relationships of equality, however. Certainly the number of human beings on earth is greater than 20. Also, the average American consumes less than 10 grams of vitamin C every day. These types of relationships are not relationships of equality, but rather, relationships of inequality .

Linear Inequalities

Linear Inequality

A linear inequality is a mathematical statement that one linear expression is greater than or less than another linear expression.

Inequality Notation

The following notation is used to express relationships of inequality:

\(>\) Strictly Greater Than

\(<\) Strictly Less Than

\(\ge\) Greater than or equal to

\(\le\) Less than or equal to

Note that the expression \(x > 12\) has infinitely many solutions. Any number strictly greater than 12 will satisfy the statement. Some solutions are \(13, 15, 90, 12.1, 16.3\) and \(102.51\).

Sample Set A

The following are linear inequalities in one variable.

Example \(\PageIndex{1}\)

1. \(x \leq 12\) 2. \(x+7>4\) 3. \(y+3 \geq 2 y-7\) 4. \(P+26<10(4 P-6)\) 5. \(\dfrac{2 r-9}{5}>15\)

The following are not linear inequalities in one variable.

Example \(\PageIndex{2}\)

1. \(x^{2}<4\) The term \(x^{2}\) is quadratic, not linear. 2. \(x \leq 5 y+3\) There are two variables. This is a linear inequality in two variables. 3. \(y+1 \neq 5\) Although the symbol \(\neq\) certainly expresses an inequality, it is customary to use only the symbols \(<,>, \leq, \geq\).

A linear equation, we know, may have exactly one solution, infinitely many solutions, or no solution. Speculate on the number of solutions of a linear inequality. ( Hint: Consider the inequalities \(x<x−6\) and \(x\ge9\).)

A linear inequality may have infinitely many solutions, or no solutions.

The Algebra of Linear Inequalities

Inequalities can be solved by basically the same methods as linear equations. There is one important exception that we will discuss in item 3 of the algebra of linear inequalities.

Let \(a, b\), and \(c\) represent real numbers and assume that:

\(a < b\) (or \(a > b\))

Then, if \(a < b\):

• \(a+c < b+c\) and \(a-c < b-c\). If any real number is added to or subtracted from both sides of an inequality, the sense of the inequality remains unchanged.
• If \(c\) is a positive real number, then if \(a < b\), \(ac < bc\) and \(\dfrac{a}{c} < \dfrac{b}{c}\). If both sides of an inequality are multiplied or divided by the same positive number the sense of the inequality remains unchanged.
• If \(c\) is a negative real number, then if \(a < b\), \(ac > bc\) and \(\dfrac{a}{c} > \dfrac{b}{c}\) If both sides of an inequality are multiplied or divided by the same negative number, the inequality sign must be reversed (change direction) in order for the resulting inequality to be equivalent to the original inequality. (See problem 4 in the next set of examples.)

For example, consider the inequality \(3 < 7\).

Example \(\PageIndex{3}\)

For \(3<7,\) if 8 is added to both sides, we get \(3+8<7+8\) \( 11<15 \) True

Example \(\PageIndex{4}\)

For \(3 < 7\), if 8 is subtracted from both sides, we get:

\(3-8 < 7-8\)

\(-5 < -1\)

Example \(\PageIndex{5}\)

For \(3 < 7\), if both sides are multiplied by 8 (a positive number), we get:

\(8(3) > 8(7)\)

\(24 < 56\)

Example \(\PageIndex{6}\)

For \(3<7\), if both sides are multiplied by −8 (a negative number), we get

\((−8)3>(−8)7\)

Notice the change in direction of the inequality sign.

\(−24>−56\)

If we had forgotten to reverse the direction of the inequality sign we would have obtained the incorrect statement \(−24<−56\).

Example \(\PageIndex{7}\)

For \(3<7\), if both sides are divided by 8 (a positive number), we get

\(\dfrac{3}{8} < \dfrac{7}{8}\)

Example \(\PageIndex{8}\)

For \(3 < 7\), if both sides are divided by -8 (a negative number), we get:

\(\dfrac{3}{-8} > frac{7}{-8}\)

True, since \(-.375 > .875\)

Sample Set B

Solve the following linear inequalities. Draw a number line and place a point at each solution.

Example \(\PageIndex{9}\)

\(3x > 15\) Divide both sides by 3. The 3 is a positive number, so we need not reverse the sense of the inequality.

\(x > 5\)

Thus, all numbers strictly greater than 5 are solutions to the inequality \(3x > 15\)

Example \(\PageIndex{10}\)

\(2y-1 \le 16\) Add 1 to both sides.

\(2y \le 17\) Divide both sides 2.

\(y \le \dfrac{17}{2}\)

Example \(\PageIndex{11}\)

\(-8x + 5 < 14\) Subtract 5 both from both sides.

\(-8x < 9\) Divide both sides by -8. We must reverse the sense of the inequality since we are divide by a neagtive number.

\(x > -\dfrac{9}{8}\)

Example \(\PageIndex{12}\)

\(5-3(y+2) < 6y - 10\) \(5-3y-6 < 6y-10\) \(-3y-1 < 6y-10\) \(-9y < -9\) \(y > 1\)

Example \(\PageIndex{13}\)

\(\dfrac{2z+7}{-4} \ge -6\) Multiply by -4 \(2x+7 \le 24\) Notice the change in the sense of the inequality. \(2z \le 17\) \(z \le \dfrac{17}{2}\)

Practice Set B

Solve the following linear inequalities.

Practice Problem \(\PageIndex{1}\)

\(y−6≤5\)

\(y≤11\)

Practice Problem \(\PageIndex{2}\)

\(x+4>9\)

Practice Problem \(\PageIndex{3}\)

\(4x−1≥15\)

Practice Problem \(\PageIndex{4}\)

\(−5y+16≤7\)

\(y \ge \dfrac{9}{5}\)

Practice Problem \(\PageIndex{5}\)

\(7(4s−3)<2s+8\)

\(s < \dfrac{29}{2}\)

Practice Problem \(\PageIndex{6}\)

\(5(1−4h)+4<(1−h)2+6\)

\(h > \dfrac{1}{18}\)

Practice Problem \(\PageIndex{7}\)

\(18≥4(2x−3)−9x\)

\(x≥−30\)

Practice Problem \(\PageIndex{8}\)

\(-\dfrac{3b}{16} \le 4\)

\(b \ge \dfrac{-64}{3}\)

Practice Problem \(\PageIndex{9}\)

\(\dfrac{-7z+10}{-12} < -1\)

\(z < -\dfrac{2}{7}\)

Practice Problem \(\PageIndex{10}\)

\(-x -\dfrac{2}{3} \le \dfrac{5}{6}\)

\(x \ge \dfrac{-3}{2}

Compound Inequalities

Compound Inequality

Another type of inequality is the compound inequality . A compound inequality is of the form:

\(a < x < b\)

There are actually two statements here. The first statement is \(a<x\). The next statement is \(x<b\). When we read this statement we say "\(a\) is less than \(x\)," then continue saying "and \(x\) is less than \(b\)."

Just by looking at the inequality we can see that the number \(x\) is between the numbers \(a\) and \(b\). The compound inequality \(a<x<b\) indicates "betweenness." Without changing the meaning, the statement \(a<x\) can be read \(x>a\). (Surely, if the number \(a\) is less than the number \(x\), the number \(x\) must be greater than the number \(a\).) Thus, we can read \(a<x<b\) as "\(x\) is greater than a and at the same time is less than \(b\)." For example:

\(4 < x < 9\).

The letter \(x\) is some number strictly between \(4\) and \(9\). Hence, \(x\) is greater than \(4\) and, at the same time, less than \(9\). The numbers \(4\) and \(9\) are not included so we use open circles at these points.

\(-2 < z < 0\).

The \(z\) stands for some number between \(-2\) and \(0\). Hence, \(z\) is greater than \(-2\) but also less than \(0\).

\(1 < x + 6 < 8\).

The expression \(x + 6\) represents some number strictly between \(1\) and \(8\). Hence, \(x + 6\) represents some number strictly greater than \(1\), but less than \(8\).

\(\dfrac{1}{4} \le \dfrac{5x-2}{6} \le \dfrac{7}{9}\).

The term \(\dfrac{5x-2}{6}\) represents some number between and including \(\dfrac{1}{4}\) and \(\dfrac{7}{9}\). Hence, \(\dfrac{5x-2}{6}\) represents some number greater than or equal to \(\dfrac{1}{4}\) but less than or equal to \(\dfrac{7}{9}\).

Consider problem 3 above, \(1<x+6<8\). The statement says that the quantity \(x+6\) is between \(1\) and \(8\). This statement will be true for only certain values of \(x\). For example, if \(x=1\), the statement is true since \(1<1+6<8\). However, if \(x=4.9\), the statement is false since \(1<4.9+6<8\) is clearly not true. The first of the inequalities is satisfied since \(1\) is less than \(10.9\), but the second inequality is not satisfied since \(10.9\) is not less than \(8\).

We would like to know for exactly which values of \(x\) the statement \(1<x+6<8\) is true. We proceed by using the properties discussed earlier in this section, but now we must apply the rules to all three parts rather than just the two parts in a regular inequality.

Sample Set C

Example \(\pageindex{14}\).

Solve \(1 < x + 6 < 8\).

\(1−6<x+6−6<8−6\) Subtract \(6\) from all three parts.

\(-5 < x < 2\)

Thus, if \(x\) is any number strictly between \(-5\) and \(2\), the statement \(1 < x+6 < 8\) will be true.

Solve \(-3 < \dfrac{-2x-7}{5} < 8\)

\(-3 < \dfrac{-2x-7}{5}(5) < 8(5)\) Multiply each part by \(5\).

\(-15 < -2x-7 < 40\). Add \(7\) to all three parts.

\(-8 < -2x < 47\) Divide all three parts by \(-2\).

\(4 > x > -\dfrac{47}{2}\) Remember to reverse the direction of the inequality signs.

\(-\dfrac{47}{2} < x < 4\). It is customary (but not necessary) to write the inequality so that inequality arrows point to the left.

Thus, if \(x\) is any number between \(-\dfrac{47}{2}\) and \(4\), the original inequality will be satisfied.

Practice Set C

Find the values of x that satisfy the given continued inequality.

Practice Problem \(\PageIndex{11}\)

\(4<x−5<12\)

\(9<x<17\)

\(−3<7y+1<18\)

\(-\dfrac{4}{7} < y < \dfrac{17}{7}\)

\(0≤1−6x≤7\)

\(-1 \le x \le \dfrac{1}{6}\)

\(-5 \le \dfrac{2x+1}{3} \le 10\)

\(-8 \le x \le \dfrac{29}{2}\)

\(9 < \dfrac{-4x+5}{-2} < 14\)

\(\dfrac{23}{4} < x < \dfrac{33}{4}\)

Does \(4<x<−1\) have a solution?

For the following problems, solve the inequalities.

Exercise \(\PageIndex{1}\)

\(x+7<12\)

Exercise \(\PageIndex{2}\)

\(y−5≤8\)

Exercise \(\PageIndex{3}\)

\(y+19≥2\)

\(y≥−17\)

Exercise \(\PageIndex{4}\)

\(x−5>16\)

Exercise \(\PageIndex{5}\)

\(3x−7≤8\)

Exercise \(\PageIndex{6}\)

\(9y−12≤6\)

Exercise \(\PageIndex{7}\)

\(2z+8<7\)

\(z < -\dfrac{1}{2}\)

Exercise \(\PageIndex{8}\)

\(4x−14>21\)

Exercise \(\PageIndex{9}\)

\(−5x≤20\)

\(x≥−4\)

Exercise \(\PageIndex{10}\)

\(−8x<40\)

Exercise \(\PageIndex{11}\)

\(−7z<77\)

\(z>−11\)

Exercise \(\PageIndex{12}\)

\(−3y>39\)

Exercise \(\PageIndex{13}\)

\(\dfrac{x}{4} \ge 12\)

\(x≥48\)

Exercise \(\PageIndex{14}\)

\(\dfrac{y}{7} > 3\)

Exercise \(\PageIndex{15}\)

\(\dfrac{2x}{9} \ge 4\)

\(x≥18\)

Exercise \(\PageIndex{16}\)

\(\dfrac{5y}{2} \ge 15\)

Exercise \(\PageIndex{17}\)

\(\dfrac{10x}{3} \le 4\)

\(x \le \dfrac{6}{5}\)

Exercise \(\PageIndex{18}\)

\(-\dfrac{5y}{4} < 8\)

Exercise \(\PageIndex{19}\)

\(\dfrac{-12b}{5} < 24\)

\(b>−10\)

Exercise \(\PageIndex{20}\)

\(\dfrac{-6a}{7} \le -24\)

Exercise \(\PageIndex{21}\)

\(\dfrac{8x}{-5} > 6\)

\(x < -\dfrac{15}{4}\)

Exercise \(\PageIndex{22}\)

\(\dfrac{14y}{-3} \ge -18\)

Exercise \(\PageIndex{23}\)

\(\dfrac{21y}{-8} < -2\)

\(y < \dfrac{16}{21}\)

Exercise \(\PageIndex{24}\)

\(−3x+7≤−5\)

Exercise \(\PageIndex{25}\)

\(−7y+10≤−4\)

Exercise \(\PageIndex{26}\)

\(6x−11<31\)

Exercise \(\PageIndex{27}\)

\(3x−15≤30\)

\(x≤15\)

Exercise \(\PageIndex{28}\)

\(-2y + \dfrac{4}{3} \le -\dfrac{2}{3}\)

Exercise \(\PageIndex{29}\)

\(5(2x−5)≥15\)

Exercise \(\PageIndex{30}\)

\(4(x+1)>−12\)

Exercise \(\PageIndex{31}\)

\(6(3x−7)≥48\)

Exercise \(\PageIndex{32}\)

\(3(−x+3)>−27\)

Exercise \(\PageIndex{33}\)

\(−4(y+3)>0\)

\(y<−3\)

Exercise \(\PageIndex{34}\)

\(−7(x−77)≤0\)

Exercise \(\PageIndex{35}\)

\(2x−1<x+5\)

Exercise \(\PageIndex{36}\)

\(6y+12≤5y−1\)

Exercise \(\PageIndex{37}\)

\(3x+2≤2x−5\)

\(x≤−7\)

Exercise \(\PageIndex{38}\)

\(4x+5>5x−11\)

Exercise \(\PageIndex{39}\)

\(3x−12≥7x+4\)

\(x≤−4\)

Exercise \(\PageIndex{40}\)

\(−2x−7>5x\)

Exercise \(\PageIndex{41}\)

\(−x−4>−3x+12\)

Exercise \(\PageIndex{42}\)

\(3−x≥4\)

Exercise \(\PageIndex{43}\)

\(5−y≤14\)

\(y≥−9\)

Exercise \(\PageIndex{44}\)

\(2−4x≤−3+x\)

Exercise \(\PageIndex{45}\)

\(3[4+5(x+1)]<−3\)

\(x<−2\)

Exercise \(\PageIndex{46}\)

\(2[6+2(3x−7)]≥4\)

Exercise \(\PageIndex{47}\)

\(7[−3−4(x−1)]≤91\)

\(x≥−3\)

Exercise \(\PageIndex{48}\)

\(−2(4x−1)<3(5x+8)\)

Exercise \(\PageIndex{49}\)

\(−5(3x−2)>−3(−x−15)+1\)

Exercise \(\PageIndex{50}\)

\(−.0091x≥2.885x−12.014\)

Exercise \(\PageIndex{51}\)

What numbers satisfy the condition: twice a number plus one is greater than negative three?

\(x>−2\)

Exercise \(\PageIndex{52}\)

What numbers satisfy the condition: eight more than three times a number is less than or equal to fourteen?

Exercise \(\PageIndex{53}\)

One number is five times larger than another number. The difference between these two numbers is less than twenty-four. What are the largest possible values for the two numbers? Is there a smallest possible value for either number?

First number: any number strictly smaller that 6. Second number: any number strictly smaller than 30. No smallest possible value for either number. No largest possible value for either number.

Exercise \(\PageIndex{54}\)

The area of a rectangle is found by multiplying the length of the rectangle by the width of the rectangle. If the length of a rectangle is 8 feet, what is the largest possible measure for the width if it must be an integer (positive whole number) and the area must be less than 48 square feet?

Exercises for Review

Exercise \(\pageindex{55}\).

Simplify\((x^2y^3z^2)^5\)

\(x^{10}y^{15}z^{10}\)

Exercise \(\PageIndex{56}\)

Simplify \(−[−(−|−8|)]\).

Exercise \(\PageIndex{57}\)

Find the product. \((2x−7) (x+4)\).

\(2x^2 + x - 28\)

Exercise \(\PageIndex{58}\)

Twenty-five percent of a number is \(12.32\). What is the number?

Exercise \(\PageIndex{59}\)

The perimeter of a triangle is 40 inches. If the length of each of the two legs is exactly twice the length of the base, how long is each leg?

Chapter 4: Linear Equations and Inequalities in One Variable

4.2.1 an introduction to inequalities, learning outcomes.

• Represent inequalities on a number line
• Represent inequalities using interval notation
• Describe solutions to inequalities
• Inequality : a mathematical statement that compares two expressions using the ideas of greater than or less than
• Real number line : a line that represents all real numbers from negative infinity to infinity
• Interval notation : sets with parentheses representing the strict inequalities < or > and or brackets representing the inequalities ≤ or ≥
• Solution set : the set formed by all solutions of an inequality
• Set-builder notation : defines the variable and uses an inequality inside a pair of braces

Inequalities

An inequality is a mathematical statement that compares two expressions using the ideas of greater than or less than. We use the following symbols with inequalities.

When we have inequalities that involve variables, there is often an infinite number of values the variable can take.

For example:

• [latex]{x}\lt{9}[/latex] indicates the list of numbers that are less than [latex]9[/latex].
• [latex]-5\le{t}[/latex] indicates all the numbers that are greater than or equal to [latex]-5[/latex].

Note how placing the variable on the left or right of the inequality sign can change whether you are looking for greater than or less than.

• [latex]x\lt5[/latex] means all the real numbers that are less than 5, whereas;
• [latex]5\lt{x}[/latex] means that 5 is less than x, or we could rewrite this with the x on the left: [latex]x\gt{5}[/latex]. Note how the inequality is still pointing the same direction relative to x. This statement represents all the real numbers that are greater than 5, which is easier to interpret than 5 is less than x.

Representing Inequalities on a Number Line

As we saw in 1.1.2, inequalities can also be graphed on a number line. Below are three examples of inequalities and their graphs.  Graphs are a very helpful way to visualize information, especially when that information represents an infinite list of numbers.

[latex]x\leq -4[/latex]. This translates to all the real numbers on a number line that are less than or equal to [latex]4[/latex].

[latex]{x}\geq{-3}[/latex]. This translates to all the real numbers on the number line that are greater than or equal to -3.

Each of these graphs begins with a circle—either an open or closed (shaded) circle. This point is often called the end point of the solution. A closed, or shaded, circle is used to represent the inequalities greater than or equal to  [latex] \displaystyle \left(\geq\right) [/latex] or less than or equal to  [latex] \displaystyle \left(\leq\right) [/latex]. The point is part of the solution. An open circle is used for greater than (>) or less than (<). The point is not part of the solution.

The graph then extends endlessly in one direction. This is shown by a line with an arrow at the end. For example, notice that for the graph of [latex] \displaystyle x\geq -3[/latex] shown above, the end point is [latex]−3[/latex], represented with a closed circle since the inequality is greater than or equal to [latex]−3[/latex]. The blue line is drawn to the right on the number line because the values in this area are greater than [latex]−3[/latex]. The arrow at the end indicates that the solutions continue infinitely.

Graph the inequality [latex]x\ge 4[/latex]

We can use a number line as shown. Because the values for [latex]x[/latex] include [latex]4[/latex], we place a solid dot on the number line at [latex]4[/latex].

Then we draw a line that begins at [latex]x=4[/latex] and, as indicated by the arrowhead, continues to positive infinity, which illustrates that the solution set includes all real numbers greater than or equal to [latex]4[/latex].

Write an inequality describing all the real numbers on the number line that are less than [latex]2[/latex]. Then draw the corresponding graph.

We need to start from the left and work right, so we start from negative infinity and end at [latex]2[/latex]. We will not include either because infinity is not a number, and the inequality does not include [latex]2[/latex].

Inequality: [latex]x\lt2[/latex]

To draw the graph, place an open dot on the number line first, and then draw a line extending to the left. Draw an arrow at the leftmost point of the line to indicate that it continues for infinity.

Interval Notation

Another commonly used, and arguably the most concise, method for describing inequalities and solutions to inequalities is called   interval notation .  We introduced this convention in 1.1.2. when we built sets with parentheses representing the strict inequalities < or > and or brackets representing the inequalities ≤ or ≥.  We also use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” For example, [latex]x \in [6,\,\infty)[/latex] represents all x-values that are greater than or equal to 6.

A few examples of an interval , or a set of numbers in which a solution falls, are [latex]\left[-2,6\right)[/latex], or all numbers between [latex]-2[/latex] and [latex]6[/latex], including [latex]-2[/latex], but not including [latex]6[/latex]; [latex]\left(-1,0\right)[/latex], all real numbers between, but not including [latex]-1[/latex] and [latex]0[/latex]; and [latex]\left(-\infty,1\right][/latex], all real numbers less than and including [latex]1[/latex]. The table below outlines the possibilities. Remember to read inequalities from left to right, just like text.

The table below describes all the possible inequalities that can occur and how to write them using interval notation, where a and b are real numbers.

Describe the inequality [latex]x\ge 4[/latex] using interval notation Solution

The solutions to [latex]x\ge 4[/latex] are represented as [latex]\left[4,\infty \right)[/latex].

Note the use of a bracket on the left because 4 is included in the solution set.

Use interval notation to indicate all real numbers greater than or equal to [latex]-2[/latex].

Use a bracket on the left of [latex]-2[/latex] and parentheses after infinity: [latex]\left[-2,\infty \right)[/latex]. The bracket indicates that [latex]-2[/latex] is included in the set with all real numbers greater than [latex]-2[/latex] to infinity.

The following video shows examples of how to write inequalities in the three ways presented here: as an inequality, in interval notation, and with a graph.

Set-Builder Notation

The solutions to inequalities form a set of numbers, so set-builder notation (see section 1.1.2) can also be used to represent the solution set . Set-builder notation defines the variable and uses an inequality inside a pair of braces. For example, the solution set [latex]\left ( 4,\,\infty \right )[/latex] is written in set-builder notation as [latex]\{ x\;\large |\;\normalsize x\gt 4,\;x\in\mathbb{R} \; \}[/latex]. This is read the set of all [latex]x[/latex]-values such that [latex]x[/latex] is a real number greater than 4.

Convert the solution set from interval notation to set-builder notation:  [latex]x\in \left (-\infty,\; 7\right ][/latex]

[latex]x[/latex] lies between [latex]-\infty[/latex] and [latex]7[/latex], so [latex]x[/latex] is less than or equal to [latex]7[/latex].

[latex]\{\;x\;\large|\;\;x\le 7,\;\;x\in\mathbb{R}\}[/latex]

Convert the solution set from set-builder notation to interval notation: [latex]\{\;x\;\large|\;\;x\ge -2,\;\;x\in\mathbb{R}\}[/latex]

[latex]x[/latex] is greater than or equal to [latex]-2[/latex] so the interval starts at [latex]-2[/latex] and goes to [latex]\infty[/latex].

[latex]x\in\left [ -2,\;\infty\right )[/latex]

1. Convert the solution set from interval notation to set-builder notation:  [latex]x\in \left (-3,\;\infty\right )[/latex]

2. Convert the solution set from set-builder notation to interval notation: [latex]\{\;x\;\large|\;\;x\le 8,\;\;x\in\mathbb{R}\}[/latex]

• [latex]\{\;x\;\large|\;\;x\gt -3,\;\;x\in\mathbb{R}\}[/latex]
• [latex]x\in \left (-\infty,\;8\right ][/latex]
• Graph Linear Inequalities in One Variable (Basic). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/-kiAeGbSe5c . License : CC BY: Attribution
• Set-Builder Notation. Authored by : Hazel McKenna. Provided by : Utah Valley University. License : CC BY: Attribution
• Graph Basic Inequalities and Express Using Interval Notation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/X0xrHKgbDT0 . License : CC BY: Attribution
• Revised and adapted: College Algebra. Authored by : Jay Abramson, et al.. Provided by : Lumen Learning. Located at : . License : CC BY: Attribution
• Revised and adapted: Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution
• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution

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Solving Inequalities in One Variable

7th - 9th grade, mathematics.

12 questions

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Solve the inequality:

–2x > –10

3(x – 1) ≥ –9

Graph the solution set of the inequality.

–4n + 1 > –23

2x + 5 > 2x + 8

Infinite Solutions

No Solution

3(n + 1) ≤ 3n + 4

2n + 4 < 16

4n – 5 < 4(n – 2) + 9

10 – 2x ≥ –4

–3x + 1 ≤ 13

Solve the compound inequality.

–10 ≤ 2x + 4 < 22

–3 < x ≤ 13

–7 ≤ x < 9

–7 < x ≤ 9

–3 ≤ x < 13

Graph the inequality:

x ≤ 5 or x > 9

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7.2 Inequalities in One Variable

When learning about domain and range, you learned about inequalities and using set-builder and interval notation to represent them. In this section, we will explore how to solve linear and absolute value inequalities in one variable. The process is very similar to solve equations, but instead of the solution being a single value, the solution will be an inequality.

Notice that if an inequality is true, like 2 < 5, then these operations result in a true statement as well, just like with equations:

Adding a number to both sides:

[latex]\begin{align*} 2+4&<5+4&\\ 6&<9 & True\\ \end{align*}[/latex]

Subtracting a number on both sides:

[latex]\begin{align*} 2 – 3 &< 5 – 3&\\ -1 &< 2 & True\\ \end{align*}[/latex]

Multiplying a positive number on both sides:

[latex]\begin{align*} 2(3) &< 5(3)\\ 6&<15 & True\\ \end{align*}[/latex]

Dividing by a positive number on both sides:

[latex]\begin{align*} \frac{2}{2}&<\frac{5}{2}\\ 1&<2.5 & True\\ \end{align*}[/latex]

We can use these operations just like when solving equations.

Example 7.2A

Solve [latex]3x + 7 \geq 1[/latex]

[latex]\begin{align*} 3x+7&\geq1 & {\textrm{Subtract 7 from both sides}}\\ 3x&\geq-6 & {\textrm{Divide both sides by 3}}\\ x&\geq-2 \end{align*}[/latex]

This inequality represents the solution set. It tells us that all numbers greater than or equal to -2 will satisfy the original inequality. We could also write this solution in interval notation, as [latex][-2, \infty)[/latex]

To understand what is happening, we could also consider the problem graphically. If we were to graph the equation [latex]y = 3x +7[/latex], then solving [latex]3x + 7 \geq 1[/latex] would correspond with asking “for what values of [latex]x[/latex] is [latex]y\geq 1[/latex]”.

Notice that the part of the graph where this is true corresponds to where [latex]x\geq -2[/latex].

While most operations in solving inequalities are the same as in solving equations, we run into a problem when multiplying or dividing both sides by a negative number. Notice, for example:

[latex]\begin{align*} 2(-3)&<5(-3)\\ -6&<-15 & {\textrm{Not True}}\\ \end{align*}[/latex]

To account for this, when multiplying or dividing by a negative number, we must reverse the sign of the inequality.

Rules for Solving Linear Inequalities

• You may add or subtract a positive or negative number to both sides of the inequality.
• You can multiply or divide both sides of the inequality by a positive number.
• You can multiply or divide both sides of the inequality by a negative number, but you must reverse the direction of the inequality.

Example 7.2B

Solve [latex]12 - 4x < 6[/latex]

[latex]\begin{align*} 12-4x& \frac{-6}{-4} &{\text{Simplify}}\\ x&=\frac{3}{2}\\ \end{align*}[/latex]

Give It Some Thought

1. Solve: [latex]6 + 2x\leq18 + 5x[/latex]

Example 7.2 C

A company spends $1200 per day on overhead and labor, and each item they produce costs $5 for materials. If they sell the items for $10 each, how many items will they need to sell each day for their profits to be positive?

Answer: We want the profit to be positive: [latex]P>0[/latex].

Costs: [latex]C(q) = 1200 + 5q[/latex]

Revenue: [latex]R(q) = 10q[/latex]

Profit: [latex]P(q) = 10q – (1200 + 5q) = 5q – 1200[/latex]

Solving [latex]P(q) > 0[/latex]:

[latex]\begin{align*} 5q&-1200>0\\ 5q&>1200\\ q&>240\\ \end{align*}[/latex]

The company will need to less at least 240 items a day to make a profit.

Compound Inequalities

• Break it into two separate inequalities
• Solve each inequality separately
• Combine the solutions if possible.

Example 7.2D

Solve [latex]-1 < -3x + 5 < 14[/latex].

[latex]\begin{align*} -6& x\\ \end{align*}[/latex]

[latex]\begin{align*} -3x& -3\\ \end{align*}[/latex]

Now we can combine these solution sets. The numbers where both [latex]2 > x[/latex] and [latex]x > -3[/latex] are true is the set: [latex]2 > x > -3[/latex]

With this particular inequality, it would also be possible to skip the step of breaking it apart, and instead just subtract 5 from all three “parts” of the inequality. This works for simple problems like this, but may fail if the inequality has variables in more than one “part” of the inequality.

2. Solve: [latex]4\leq2x + 6 < 16[/latex]

Absolute Value

So far in this section we have been looking at inequalities that are linear. We will now turn to absolute value inequalities. The absolute value function is a piecewise-defined function made up of two linear functions.

Absolute Value Function

The absolute value function can be defined as

[latex]\begin{align*} f(x) = | x | &=&\\ &&{x} &&{x \geq 0}\\ &&{- x} &&{x < 0}\\ \end{align*}[/latex]

The graph of the absolute value looks like a V:

The absolute value function is commonly used to determine the distance between two numbers on the number line. Given two values [latex]a[/latex] and [latex]b[/latex], then [latex]|a - b|[/latex] will give the distance, a positive quantity, between these values, regardless of which value is larger.

Example 7.2E

Describe all values, [latex]x[/latex], within a distance of 4 from the number 5.

Answer: We want the distance between [latex]x[/latex] and 5 to be less than or equal to 4. The distance can be represented using the absolute value, giving the expression

[latex]|x - 5|\leq4[/latex]

Example 7.2F

A 2010 poll reported 78% of Americans believe that people who are gay should be able to serve in the US military, with a reported margin of error of 3% [1] . The margin of error tells us how far off the actual value could be from the survey value [2] . Express the set of possible values using absolute values.

Answer: Since we want the size of the difference between the actual percentage, [latex]p[/latex], and the reported percentage to be less than 3%,[latex]|p - 78|\leq3[/latex]

3. Students who score within 20 points of 80 will pass the test. Write this as a distance from 80 using the absolute value notation.

Solving Absolute Value Equations

To solve an equation like [latex]8 = |2x - 6|[/latex], we can notice that the absolute value will be equal to 8 if the quantity inside the absolute value were 8 or -8. This leads to two different equations we can solve independently:

[latex]\begin{align*} 2x-6&=8\\ 2x&=14\\ x&=7\\ \end{align*}[/latex]

[latex]\begin{align*} 2x-6&=-8\\ 2x&=-2\\ x&=-1\\ \end{align*}[/latex]

Solutions to Absolute Value Equations

An equation of the form [latex]|A| = B[/latex], with [latex]B\geq 0[/latex], will have solutions when [latex]A = B[/latex] or [latex]A = -B[/latex].

Example 7.2G

Solve: [latex]0 = |4x + 1| - 7[/latex]

Answer: Isolate the absolute value on one side of the equation}

[latex]\begin{align*} 0&=|4x+1|-7\\ 7&=|4x+1| \end{align*}[/latex]

Now we can break this into two separate equations:

[latex]\begin{align*} 7 &= 4x + 1\\ 6 &= 4x\\ x &= \frac{6}{4}\\ &= \frac{3}{2} \end{align*}[/latex]

[latex]\begin{align*} - 7 &= 4x + 1\\ - 8 &= 4x\\ x &= \frac{{ - 8}}{4}\\ &= - 2\\ \end{align*}[/latex]

There are two solutions: [latex]x = \frac{3}{2}[/latex] and [latex]x = -2[/latex]

Example 7.2H

Solve [latex]1 = 4|x - 2| + 2[/latex]

Answer: Isolating the absolute value on one side the equation,

[latex]\begin{align*} 1 &= 4|x - 2| + 2\\ -1 &= 4|x - 2|\\ -\frac{1}{4} &= |x - 2|\\ \end{align*}[/latex]

At this point, we notice that this equation has no solutions – the absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value.

4. Find the horizontal & vertical intercepts for the function

[latex]f(x) = -|x + 2| + 3[/latex]

Solving Absolute Value Inequalities

When absolute value inequalities are written to describe a set of values, like the inequality [latex]|x - 5|\leq4[/latex] we wrote earlier, it is sometimes desirable to express this set of values without the absolute value, either using inequalities, or using interval notation.

We will explore two approaches to solving absolute value inequalities:

• Using a graph
• Using test values

Example 7.2I

Solve [latex]|x - 5|\leq4[/latex]

Answer: With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where [latex]|x -5| = 4[/latex]. We do this because the absolute value is a nice friendly function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve [latex]|x - 5| = 4[/latex],

[latex]\begin{align*} x - 5 &= 4\\ x &= 9 \end{align*}[/latex]

[latex]\begin{align*} x - 5 &= - 4\\ x &= 1\\ \end{align*}[/latex]

To use a graph, we can sketch the function [latex]f(x) = |x - 5|[/latex]. To help us see where the outputs are 4, the line [latex]g(x) = 4[/latex] could also be sketched.

On the graph, we can see that indeed the output values of the absolute value are equal to 4 at [latex]x = 1[/latex] and [latex]x = 9[/latex]. Based on the shape of the graph, we can determine the absolute value is less than or equal to 4 between these two points, when [latex]1\leq x\leq 9[/latex]. In interval notation, this would be the interval [latex][1, 9][/latex].

As an alternative to graphing, after determining that the absolute value is equal to 4 at [latex]x = 1[/latex] and [latex]x = 9[/latex], we know the graph can only change from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals: [latex]x 9[/latex]. To determine when the function is less than 4, we could pick a value in each interval and see if the output is less than or greater than 4.

Example 7.2J

Given the function [latex]f(x) = - \frac{1}{2}| {4x - 5} | + 3[/latex], determine for what [latex]x[/latex] values the function values are negative.

[latex]\begin{align*} -\frac{1}{2}|4x-5|&<-3\\ \end{align*}[/latex]

when we multiply both sides by -2, it reverses the inequality

[latex]\begin{align*} |4x-5| &> 6\\ \end{align*}[/latex]

Next we solve for the equality [latex]|4x - 5| = 6[/latex]

[latex]\begin{align*} 4x - 5 &= 6\\ 4x &= 11\\ x &= \frac{{11}}{4}\\ \end{align*}[/latex]

[latex]\begin{align*} 4x - 5 &= - 6\\ 4x &= - 1\\ x &= \frac{{ - 1}}{4}\\ \end{align*}[/latex]

We can now either pick test values or sketch a graph of the function to determine on which intervals the original function value are negative. Notice that it is not even really important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at [latex]x=\frac{-1}{4}[/latex] and [latex]x=\frac{11}{4}[/latex], and that the graph has been flipped.

From the graph of the function, we can see the function values are negative to the left of the first horizontal intercept at [latex]x=\frac{-1}{4}[/latex], and negative to the right of the second intercept at [latex]x=\frac{11}{4}[/latex]. This gives us the solution to the inequality:

[latex]\begin{align*} x \frac{11}{4} \end{align*}[/latex]

In interval notation, this would be [latex]( { - \infty ,\frac{{ - 1}}{4}} ) \cup ( {\frac{{11}}{4},\infty } )[/latex]

6. Solve [latex]-2|k - 4|\leq-6[/latex]

There is a third approach to solving absolute value inequalities that is formulaic. While it works, and you are welcome to use it, it is much more likely that you will remember the other approaches.

Solutions to Absolute Value Inequalities

[latex]\begin{align*} {\textrm{ To solve }}|A| B {\textrm{, solve: }}A > B{\textrm{ or }}A < -B\\ \end{align*}[/latex]

Example 7.2 K

Solve [latex]3|x + 4| - 2\geq7[/latex]

Answer: We need to start by isolating the absolute value:

[latex]\begin{align*} 3|x+4|-2&\geq7 &{\textrm{Add 2 to both sides}}\\ 3|x+4|&\geq9 &{\textrm{Divide both sides by 3}}\\ |x+4|&\geq3\\ \end{align*}[/latex]

Now we can break this apart and solve each piece separately:

[latex]\begin{align*} x + 4&\geq 3\\ x&\geq-1\\ \end{align*}[/latex]

[latex]\begin{align*} x + 4&\leq-3\\ x&\leq-7\\ \end{align*}[/latex]

In interval notation, this would be [latex]( - \infty , - 7] \cup [ - 1,\infty )[/latex]

Give It Some Thought Answers

• [latex]x\geq-4[/latex]
• [latex]4\leq wx+6<16[/latex]
• Using the variable [latex]p[/latex], for passing, [latex]|p-80|\leq 20[/latex]
• [latex]f(x)=-|x+2|+3[/latex]
• [latex]f(0) = 1[/latex], so the vertical intercept is at [latex](0, 1)[/latex]. [latex]f(x)= 0[/latex] when [latex]x = -5[/latex] and [latex]x = 1[/latex] so the horizontal intercepts are at [latex](-5, 0)[/latex] and [latex](1, 0)[/latex].
• [latex]k 7[/latex]; in interval notation this would be [latex]( { - \infty ,1}) \cup( {7,\infty })[/latex]

Section Exercises

Work on section 7.2 exercises in Fundamentals of Business Math Exercises . Discuss your solutions with your peers and/or course instructor.

You may consult answers to select exercises: Fundamentals of Business Math Exercises – Select Answers

• http://www.pollingreport.com/civil.htm , retrieved August 4, 2010 ↵
• Technically, margin of error usually means that the surveyors are 95% confident that actual value falls within this range. ↵

Fundamentals of Business Math Copyright © 2021 by Ana Duff, adapted from work by J. Olivier and D. Lippman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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11.3 Systems of Nonlinear Equations and Inequalities: Two Variables

Learning objectives.

In this section, you will:

• Solve a system of nonlinear equations using substitution.
• Solve a system of nonlinear equations using elimination.
• Graph a nonlinear inequality.
• Graph a system of nonlinear inequalities.

Halley’s Comet ( Figure 1 ) orbits the sun about once every 75 years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear.

In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations.

Solving a System of Nonlinear Equations Using Substitution

A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form A x + B y + C = 0. A x + B y + C = 0. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.

Intersection of a Parabola and a Line

There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.

Possible Types of Solutions for Points of Intersection of a Parabola and a Line

Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.

• No solution. The line will never intersect the parabola.
• One solution. The line is tangent to the parabola and intersects the parabola at exactly one point.
• Two solutions. The line crosses on the inside of the parabola and intersects the parabola at two points.

Given a system of equations containing a line and a parabola, find the solution.

• Solve the linear equation for one of the variables.
• Substitute the expression obtained in step one into the parabola equation.
• Solve for the remaining variable.
• Check your solutions in both equations.

Solving a System of Nonlinear Equations Representing a Parabola and a Line

Solve the system of equations.

Solve the first equation for x x and then substitute the resulting expression into the second equation.

Expand the equation and set it equal to zero.

Solving for y y gives y = 2 y = 2 and y = 1.   y = 1.   Next, substitute each value for y y into the first equation to solve for x . x . Always substitute the value into the linear equation to check for extraneous solutions.

The solutions are ( 1 , 2 ) ( 1 , 2 ) and ( 0 , 1 ) , ( 0 , 1 ) , which can be verified by substituting these ( x , y ) ( x , y ) values into both of the original equations. See Figure 3 .

Could we have substituted values for y y into the second equation to solve for x x in Example 1 ?

Yes, but because x x is squared in the second equation this could give us extraneous solutions for x . x .

For y = 1 y = 1

This gives us the same value as in the solution.

For y = 2 y = 2

Notice that −1 −1 is an extraneous solution.

Solve the given system of equations by substitution.

Intersection of a Circle and a Line

Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.

Possible Types of Solutions for the Points of Intersection of a Circle and a Line

Figure 4 illustrates possible solution sets for a system of equations involving a circle and a line.

• No solution. The line does not intersect the circle.
• One solution. The line is tangent to the circle and intersects the circle at exactly one point.
• Two solutions. The line crosses the circle and intersects it at two points.

Given a system of equations containing a line and a circle, find the solution.

• Substitute the expression obtained in step one into the equation for the circle.

Finding the Intersection of a Circle and a Line by Substitution

Find the intersection of the given circle and the given line by substitution.

One of the equations has already been solved for y . y . We will substitute y = 3 x −5 y = 3 x −5 into the equation for the circle.

Now, we factor and solve for x . x .

Substitute the two x -values into the original linear equation to solve for y . y .

The line intersects the circle at ( 2 , 1 ) ( 2 , 1 ) and ( 1 , −2 ) , ( 1 , −2 ) , which can be verified by substituting these ( x , y ) ( x , y ) values into both of the original equations. See Figure 5 .

Solve the system of nonlinear equations.

Solving a System of Nonlinear Equations Using Elimination

We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse.

Possible Types of Solutions for the Points of Intersection of a Circle and an Ellipse

Figure 6 illustrates possible solution sets for a system of equations involving a circle and an ellipse .

• No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other.
• One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point.
• Two solutions. The circle and the ellipse intersect at two points.
• Three solutions. The circle and the ellipse intersect at three points.
• Four solutions. The circle and the ellipse intersect at four points.

Solving a System of Nonlinear Equations Representing a Circle and an Ellipse

Let’s begin by multiplying equation (1) by −3 , −3 , and adding it to equation (2).

After we add the two equations together, we solve for y . y .

Substitute y = ± 1 y = ± 1 into one of the equations and solve for x . x .

There are four solutions: ( 5 , 1 ) , ( −5 , 1 ) , ( 5 , −1 ) , and ( −5 , −1 ) . ( 5 , 1 ) , ( −5 , 1 ) , ( 5 , −1 ) , and ( −5 , −1 ) . See Figure 7 .

Find the solution set for the given system of nonlinear equations.

Graphing a Nonlinear Inequality

All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality.

Recall that when the inequality is greater than, y > a , y > a , or less than, y < a , y < a , the graph is drawn with a dashed line. When the inequality is greater than or equal to, y ≥ a , y ≥ a , or less than or equal to, y ≤ a , y ≤ a , the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade. See Figure 8 .

Given an inequality bounded by a parabola, sketch a graph.

• Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set.
• If the boundary is included in the region (the operator is ≤ ≤ or ≥ ≥ ), the parabola is graphed as a solid line.
• If the boundary is not included in the region (the operator is < or >), the parabola is graphed as a dashed line.
• Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line.
• Shade the region representing the solution set.

Graphing an Inequality for a Parabola

Graph the inequality y > x 2 + 1. y > x 2 + 1.

First, graph the corresponding equation y = x 2 + 1. y = x 2 + 1. Since y > x 2 + 1 y > x 2 + 1 has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points ( 0 , 2 ) ( 0 , 2 ) and ( 2 , 0 ) . ( 2 , 0 ) . One point is clearly inside the parabola and the other point is clearly outside.

The graph is shown in Figure 9 . We can see that the solution set consists of all points inside the parabola, but not on the graph itself.

Graphing a System of Nonlinear Inequalities

Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region .

Given a system of nonlinear inequalities, sketch a graph.

• Find the intersection points by solving the corresponding system of nonlinear equations.
• Graph the nonlinear equations.
• Find the shaded regions of each inequality.
• Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality.

Graphing a System of Inequalities

Graph the given system of inequalities.

These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable y y will be eliminated. Then we solve for x . x .

Substitute the x -values into one of the equations and solve for y . y .

The two points of intersection are ( 2 , 4 ) ( 2 , 4 ) and ( −2 , 4 ) . ( −2 , 4 ) . Notice that the equations can be rewritten as follows.

Graph each inequality. See Figure 10 . The feasible region is the region between the two equations bounded by 2 x 2 + y ≤ 12 2 x 2 + y ≤ 12 on the top and x 2 − y ≤ 0 x 2 − y ≤ 0 on the bottom.

Access these online resources for additional instruction and practice with nonlinear equations.

• Solve a System of Nonlinear Equations Using Substitution
• Solve a System of Nonlinear Equations Using Elimination

11.3 Section Exercises

Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers.

When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution?

When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region?

If you graph a revenue and cost function, explain how to determine in what regions there is profit.

If you perform your break-even analysis and there is more than one solution, explain how you would determine which x -values are profit and which are not.

For the following exercises, solve the system of nonlinear equations using substitution.

   x + y = 4 x 2 + y 2 = 9    x + y = 4 x 2 + y 2 = 9

         y = x −3 x 2 + y 2 = 9          y = x −3 x 2 + y 2 = 9

         y = x x 2 + y 2 = 9          y = x x 2 + y 2 = 9

         y = − x x 2 + y 2 = 9          y = − x x 2 + y 2 = 9

         x = 2 x 2 − y 2 = 9          x = 2 x 2 − y 2 = 9

For the following exercises, solve the system of nonlinear equations using elimination.

4 x 2 −9 y 2 = 36 4 x 2 + 9 y 2 = 36 4 x 2 −9 y 2 = 36 4 x 2 + 9 y 2 = 36

x 2 + y 2 = 25 x 2 − y 2 = 1 x 2 + y 2 = 25 x 2 − y 2 = 1

2 x 2 + 4 y 2 = 4 2 x 2 −4 y 2 = 25 x −10 2 x 2 + 4 y 2 = 4 2 x 2 −4 y 2 = 25 x −10

y 2 − x 2 = 9 3 x 2 + 2 y 2 = 8 y 2 − x 2 = 9 3 x 2 + 2 y 2 = 8

x 2 + y 2 + 1 16 = 2500 y = 2 x 2 x 2 + y 2 + 1 16 = 2500 y = 2 x 2

For the following exercises, use any method to solve the system of nonlinear equations.

−2 x 2 + y = −5     6 x − y = 9 −2 x 2 + y = −5     6 x − y = 9

− x 2 + y = 2 − x + y = 2 − x 2 + y = 2 − x + y = 2

x 2 + y 2 = 1           y = 20 x 2 −1 x 2 + y 2 = 1           y = 20 x 2 −1

x 2 + y 2 = 1           y = − x 2 x 2 + y 2 = 1           y = − x 2

2 x 3 − x 2 = y            y = 1 2 − x 2 x 3 − x 2 = y            y = 1 2 − x

9 x 2 + 25 y 2 = 225 ( x −6 ) 2 + y 2 = 1 9 x 2 + 25 y 2 = 225 ( x −6 ) 2 + y 2 = 1

x 4 − x 2 = y   x 2 + y = 0 x 4 − x 2 = y   x 2 + y = 0

2 x 3 − x 2 = y     x 2 + y = 0 2 x 3 − x 2 = y     x 2 + y = 0

For the following exercises, use any method to solve the nonlinear system.

x 2 + y 2 = 9          y = 3 − x 2 x 2 + y 2 = 9          y = 3 − x 2

x 2 − y 2 = 9           x = 3 x 2 − y 2 = 9           x = 3

x 2 − y 2 = 9           y = 3 x 2 − y 2 = 9           y = 3

x 2 − y 2 = 9     x − y = 0 x 2 − y 2 = 9     x − y = 0

− x 2 + y = 2 −4 x + y = −1 − x 2 + y = 2 −4 x + y = −1

− x 2 + y = 2          2 y = − x − x 2 + y = 2          2 y = − x

x 2 + y 2 = 25 x 2 − y 2 = 36 x 2 + y 2 = 25 x 2 − y 2 = 36

x 2 + y 2 = 1         y 2 = x 2 x 2 + y 2 = 1         y 2 = x 2

16 x 2 −9 y 2 + 144 = 0                 y 2 + x 2 = 16 16 x 2 −9 y 2 + 144 = 0                 y 2 + x 2 = 16

      3 x 2 − y 2 = 12 ( x −1 ) 2 + y 2 = 1       3 x 2 − y 2 = 12 ( x −1 ) 2 + y 2 = 1

      3 x 2 − y 2 = 12 ( x −1 ) 2 + y 2 = 4       3 x 2 − y 2 = 12 ( x −1 ) 2 + y 2 = 4

3 x 2 − y 2 = 12    x 2 + y 2 = 16 3 x 2 − y 2 = 12    x 2 + y 2 = 16

x 2 − y 2 − 6 x − 4 y − 11 = 0                    − x 2 + y 2 = 5 x 2 − y 2 − 6 x − 4 y − 11 = 0                    − x 2 + y 2 = 5

x 2 + y 2 −6 y = 7           x 2 + y = 1 x 2 + y 2 −6 y = 7           x 2 + y = 1

x 2 + y 2 = 6         x y = 1 x 2 + y 2 = 6         x y = 1

For the following exercises, graph the inequality.

x 2 + y < 9 x 2 + y < 9

x 2 + y 2 < 4 x 2 + y 2 < 4

For the following exercises, graph the system of inequalities. Label all points of intersection.

x 2 + y < 1 y > 2 x x 2 + y < 1 y > 2 x

x 2 + y < −5 y > 5 x + 10 x 2 + y < −5 y > 5 x + 10

x 2 + y 2 < 25 3 x 2 − y 2 > 12 x 2 + y 2 < 25 3 x 2 − y 2 > 12

x 2 − y 2 > −4 x 2 + y 2 < 12 x 2 − y 2 > −4 x 2 + y 2 < 12

x 2 + 3 y 2 > 16 3 x 2 − y 2 < 1 x 2 + 3 y 2 > 16 3 x 2 − y 2 < 1

y ≥ e x y ≤ ln ( x ) + 5 y ≥ e x y ≤ ln ( x ) + 5

y ≤ − log ( x ) y ≤ e x y ≤ − log ( x ) y ≤ e x

For the following exercises, find the solutions to the nonlinear equations with two variables.

4 x 2 + 1 y 2 = 24 5 x 2 − 2 y 2 + 4 = 0 4 x 2 + 1 y 2 = 24 5 x 2 − 2 y 2 + 4 = 0

6 x 2 − 1 y 2 = 8 1 x 2 − 6 y 2 = 1 8 6 x 2 − 1 y 2 = 8 1 x 2 − 6 y 2 = 1 8

x 2 − x y + y 2 −2 = 0 x + 3 y = 4 x 2 − x y + y 2 −2 = 0 x + 3 y = 4

x 2 − x y −2 y 2 −6 = 0 x 2 + y 2 = 1 x 2 − x y −2 y 2 −6 = 0 x 2 + y 2 = 1

x 2 + 4 x y −2 y 2 −6 = 0 x = y + 2 x 2 + 4 x y −2 y 2 −6 = 0 x = y + 2

For the following exercises, solve the system of inequalities. Use a calculator to graph the system to confirm the answer.

x y < 1 y > x x y < 1 y > x

x 2 + y < 3 y > 2 x x 2 + y < 3 y > 2 x

Real-World Applications

For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions.

Two numbers add up to 300. One number is twice the square of the other number. What are the numbers?

The squares of two numbers add to 360. The second number is half the value of the first number squared. What are the numbers?

A laptop company has discovered their cost and revenue functions for each day: C ( x ) = 3 x 2 −10 x + 200 C ( x ) = 3 x 2 −10 x + 200 and R ( x ) = −2 x 2 + 100 x + 50. R ( x ) = −2 x 2 + 100 x + 50. If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit.

A cell phone company has the following cost and revenue functions: C ( x ) = 8 x 2 −600 x + 21,500 C ( x ) = 8 x 2 −600 x + 21,500 and R ( x ) = −3 x 2 + 480 x . R ( x ) = −3 x 2 + 480 x . What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit.

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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Algebra and Trigonometry
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/11-3-systems-of-nonlinear-equations-and-inequalities-two-variables

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Helping with Math

One Variable Inequality

Introduction.

In Algebra , an inequality is a mathematical expression that uses the inequality symbol to indicate the relationship between two statements. Both sides of an inequality sign have different expressions. It signifies that the left-hand side expression should be more than or less than the right hand side expression, or vice versa. Literal inequalities are when the relationship between these two algebraic expressions is defined using inequality symbols .”An inequality exists when two real numbers or algebraic expressions are connected by the symbols “>”, “”,  .>  “, For instance, x>3 (x should be greater than 3)

If there is only one variable in the inequality, it is called an open sentence or one variable inequality. One variable inequality may be a linear or higher order. Precisely for further discussion, we discuss one variable linear inequality or one variable quadratic inequality. So, first we define these one variable inequalities and then their solution method.

One variable linear inequality

Definition:.

A linear inequality is a mathematical expression that compares two linear expressions and declares one to be bigger or less than the other.  The linear equations in one variable are equations that are written as ax + b = 0, where a, and b are two integers and x is a variable, and there is only one solution. 8x + 3 = 8, for particular, is a linear equation with only one variable. As a result, there is only one solution to this equation, which is x = 97/2.Here are a few good examples of linear inequalities that are all addressed throughout this section:

• 8x + 6 < 4x
• 145x + 9>16 − 2x + 1
• X < 6 

These are one variable linear inequality with variable x.  The power of x is one that’s why it is called linear inequality and due to one variable, they are linear one variable inequality.

Counter Examples

The counter examples of linear one variable inequality are as follows.

• 2x^4+5 < 3   it is one variable but not linear
• Y + 5X = 78
• X^2 – z = 7
• 2x + 14y > 234 and 
• 3y + 4x + 3xy       these are called inequalities two variables because here we used two variables x and y. so, it is also include in the counter example of linear one variable inequality.

When a real number is replaced for the variable in a linear inequality, the result is a true assertion. Linear inequalities can have an unbounded number of solutions or none at all. If there are infinitely many possibilities, graph the set on a line segment and/or use interval notation to illustrate the answer.

To calculate a one-variable inequality, convert it to an expression (a mathematical phrase with a “=” sign) and solve it. On a number line , write the solution, often known as a “boundary point.” The number line is divided into two parts at this point. For circumstances involving or, the boundary point is included in the solution; for instances involving strictly > or, it is excluded. Boundary points that are included in the solutions are indicated on the number line as a solid filled-in circle, whereas excluded solutions are shown as an open circle. After then, pick a number from each zone split by the border point and see if it is true or untrue in the initial inequality. If this is the case, then any integer in that area is a solution to the inequality. If it’s false, then there’s no number within this region that can solve the inequality.

We have to follow some rules while solving one variable linear inequality.

Solving inequalities follows a similar set of criteria as solving mathematical equation . When multiplying or dividing by a negative number, there is one exception.

To eliminate an inequity, we can:

• To both sides, add the same number.
• Reduce the same amount from both sides of the equation.
• Both sides should be multiplied by the same positive value.
• Subtract the same positive value from both sides.
• Change the sign and multiply both sides by the same negative value.
• Change the sign and divide both sides by the same negative number.

For example 

If we have 

X + 5 > -2 

If we multiply this inequality by -2 then we have 

  -2x – 10 < 4 

Compare these, multiply by -2 changed the sign of inequality. 

Now see some examples for solving one variable linear inequality

Hence, Inequalities can also be solved by dividing or multiplication both sides by a variable. To solve the inequality 5x > 3, for example, multiply both sides by 5 to get x > 35.  When we multiply or divide by a negative integer, though, something unusual happens. For example, we know that 5 is greater than 3. However, if both sides of the inequality 5>3 are multiplied by -2, we get 10>6. We know this isn’t true because -10 is smaller than -6.

When we multiply or divide an inequality by a negative number, this occurs, and we must reverse the sign to make the inequality true. To multiply 2 > 4 by -3, for example, we multiply the 2 and 4 by -3 each, then alter the sign to a < sign, resulting in 6 < 12.

Solve this equation: 6x – (x + 10) <0

We need separate the variable on one side of the inequality sign to resolve an inequality. We employ the same fundamental approaches used to solve the equation to isolate the factor. Most inequalities can be solved by adding or removing a constant from one side of the equation.

Convert the problem to an equation and solve it.

6x – x – 10 = 0

5 x – 10 = 0

5x = 10   this implies x = 10/5 so, x = 2 

Hence we split the real line into two intervals

(-∞, 2 )          (2 , ∞)

Now we will pick one test point in each test interval and check the sign of inequality then decide solution includes which interval. 

Hence interval  (-∞, 2) satisfies the solution region. So, this is the required solution.

On the number line, write the solution (boundary point).

Solution -2 -1 0   1     2

We use a filled-in dot since x = 2 is also a solution to the inequality (< ). In the initial inequality, test a number on either side of the boundary point. The section containing the data that make the inequality true should be highlighted.

4x > 24 was the original problem.

Divide both sides by 4

4x / 4 > 24 / 4 

x > 6    

x > 6 is a simplified version of this inequality.

The building of one house stands four storey’s higher than the another building. Find the number of stories in each building if the total number of stories in both buildings is 24.

We know that x = number of stories of one building which is  is higher than another and x + 8 = other building height.

(x + 8) + x = 24 

2x + 8 = 24 

2x = 24 – 8

x = 8 hence, one building contains 8+8 is 16 stories and other is 8.

Hence, in one variable, most linear equations have precisely one solution. In at least one example , a linear equation may have an unlimited number of solutions, but in that case, every real integer is a solution. The “simple” inequality we’re looking at, x <2, has an endless number of solutions, but not every real integer is one of them. 3 is not a solution, for example: 3< 2 is false.

Quadratic inequality of one variable

The exponent of the highest power to which the variable is raised in an equation is the degree of that equation in one variable. Inside one variable, the second degree equation is one in which the variable is increased to the power of two. A QUADRATIC EQUATION is a term that describes a second-degree equation. The word quadratic comes from the Latin similar assets, which means literally “square.” The squared term is the highest degree term in a quadratic equation . For instance, consider quadratic equations the form of the quadratic equation ax^2+bc+c > 0 or ax^2 +bx + c <0 

The terms with a lower degree than the second may or may not exist. The first-degree term and the constant term are lower-degree terms in a quadratic equation than the squared term. In the formula

Solution methods

We can solve the inequality algebraically as well as graphically

When solving a quadratic inequality, we must first determine the set of alternatives, or intervals, whereby the inequality holds. This is much more complicated than the linear example and may include more than one unique period. The steps outlined below can be used to solve quadratic inequalities.

Solving a Quadratic Inequality :

A Step-by-Step Guide Algebraically

Reorganize the inequality so that most of the expression’s parts, expressed as f ( x ), are on one side, with just an inequality connecting them to zero. For instance, f ( x ) 0 r f (x ) > 0. To determine the solutions to the equation, solve f ( x ) = 0 via factoring or otherwise. Choose test points for every interval such that values just under, between, and greater than the equation’s solutions are represented.  A sign chart can also be used to determine whether the intervals will be positively or negatively. Determine which intervals meet the inequality.

Summary of steps:

We can also summarize these steps in to the following 

• Use this action to solve a quadratic inequality:
• As though it were an equation, solve the inequality.
• The genuine solutions to the equation serve as boundary points for the inequality’s formulation.
• If the original inequality contains equality, make the border points solid circles; otherwise, make them open circles.
• Choose points from each of the boundary point-created zones. In the initial inequality, replace these “test points” with these “test points.”
• When a test point meets the original inequality, the region in which it is located is part of the solution.
• Show the solution graphically and in the form of a solution set.

Solve the problem (x – 3)(x + 2) > 0 

Write its inequality in the form of equation

(x – 3)(x + 2) = 0 (x – 3)(x + 2) = 0 (x – 3)(x + As a result of the zero product property, equation

Make the line of demarcation. So because original inequality did not involve equality, the dividing line points are open circles.

Choose points from the several regions that have been developed equation.

Examine the test points to see if they satisfy the initial inequality.

(x – 3)(x + 2) = 0

This implies (x – 3) = 0   or (x + 2 ) = 0

X = 3   or x = -2 

 The testes intervals are 

(-∞,-2)       and -2 , 3  , ( 3 , ∞)

Now, we check the signs in each tested interval of the inequality

Hence the nested intervals (-∞, -2)    and (3,∞) satisfy the inequality condition. So, their union is the required solution set of the inequality.

We were able to follow the technique for solving a quadratic inequality perfectly in the first example by first resolving the quadratic and then utilizing a symbol chart, but this is not always desirable or necessary. We’ll look at a quadratic inequality that can’t be factored in the next example.

To solve the inequality x^2+121 0, find all possible solutions. Write your response as a series of intervals.

There would be no need to rearrange this inequality because it was supplied to us with all terms on one side of the equation. Normally, we would try to solve f(x)= 0.where f(x)=x^2+121 to start looking for solutions. However, there are no solutions to this problem because x will always be bigger than zero for any real value of x. To put it another way, x^2+121>0 , As a result, the inequality’s left-hand side is always strictly bigger than zero, implying that x+1210 is never true.

This can be demonstrated graphically by noting that the graph of f(x)=x^2+121 don’t ever crosses the x-axis, as illustrated below. The solution is the empty set written as an interval.

As a result our answer for this inequality is empty set.

We solved quadratic inequalities with the right side of the inequality equal to zero in the previous two examples. While both sides of an inequality have nonzero expressions, the inequality must first be simplified to the point where first side becomes zero.

Hence we conclude 

An interval or a union of intervals is the solution to a quadratic inequality. We could use set difference terminology to write it as the complement of one interval when it is a union of two intervals.

The steps of solving a quadratic inequality algebraically are as follows: f(x) > 0

Rearrange the inequality so that all of the expression’s terms are on one side, for example, using an inequality connecting this to zero. F(x) = 0

Factor the inequality by determining the expression’s roots. F(x)

Using test points in each interval or a sign chart, find the intervals that meet the inequality. A graph of the function can also be drawn.

Follow the instructions below to solve a quadratic inequality graphically.

Rearrange the inequality so that all of the expression’s terms are on one side, and an inequality connects them to zero; for example.

To find the expression’s roots, factor the inequality by setting.

Using the equation’s roots, draw the graph of the equation and determine the direction of the parabola curve. If you’ve altered the initial inequality to change the sign of the x value, be especially cautious: Instead of using the original inequality, use the x coefficient in the rearranged form of the inequality to determine the curve’s shape.

Determine which intervals satisfy the inequality.

Hence, in general 

• A one-variable linear inequality has an infinite number of solutions. As a result, the answers are usually represented by a graph on a number line.
• A typical graph of a linear inequality is half of a number line, with all points to the left or right of the border value, and the shaded region showing inequality solutions.
• In the event of a strict inequality (or >), the border value is indicated by an open circle (), while in the case of a non-strict inequality (or), the border value is indicated by a solid circle (•).
• The test value method or the standard form approach can be used to determine which side of the border value to shade.

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