linear equation problem solving questions

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.

Solution: Then the other number = x + 9 Let the number be x. Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9) ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12 Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72 ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.

Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts. Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40 The two parts are 15 and 25.

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.

7. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair. Solution: The table cost $ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165.

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

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What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

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Practice Test on Word Problems on Linear Equations

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Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

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Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

When x increases, y increases twice as fast , so we need 2x
When x is 0, y is already 1. So +1 is also needed
And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

Exponents (like the 2 in x 2 )
Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

Slope: m = 2
Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

Finding the Midpoint of a Line Segment
Finding Parallel and Perpendicular Lines
Finding the Equation of a Line from 2 Points

Linear Equations

A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is a variable, A is a coefficient and B is constant. The standard form of a linear equation in two variables is of the form Ax + By = C. Here, x and y are variables, A and B are coefficients and C is a constant.

What is a Linear Equation?

An equation that has the highest degree of 1 is known as a linear equation . This means that no variable in a linear equation has a variable whose exponent is more than 1. The graph of a linear equation always forms a straight line.

Linear Equation Definition: A linear equation is an algebraic equation where each term has an exponent of 1 and when this equation is graphed, it always results in a straight line. This is the reason why it is named as a 'linear' equation.

There are linear equations in one variable and linear equations in two variables. Let us learn how to identify linear equations and non-linear equations with the help of the following examples.

Linear Equation Formula

The linear equation formula is the way of expressing a linear equation. This can be done in different ways. For example, a linear equation can be expressed in the standard form, the slope-intercept form, or the point-slope form. Now, if we take the standard form of a linear equation, let us learn the way in which it is expressed. We can see that it varies from case to case based on the number of variables and it should be remembered that the highest (and the only) degree of all variables in the equation should be 1.

Slope intercept form of a linear equation is y = mx + c (where m = slope and c = y-intercept)
Point slope form of a linear equation is y - y 1 = m(x - x 1 ) (where m = slope and (x 1 , y 1 ) is a point on the line)

Note: The slope of a linear equation is the amount by which the line is rising or falling. It is calculated by the formula rise/run . i.e., if (x 1 , y 1 ) and (x 2 , y 2 ) are any two points on a line then its slope is calculated using the formula (y 2 - y 1 )/(x 2 - x 1 ).

Linear Equations in Standard Form

The standard form or the general form of linear equations in one variable is written as, Ax + B = 0; where A and B are real numbers , and x is the single variable. The standard form of linear equations in two variables is expressed as, Ax + By = C; where A, B and C are any real numbers, and x and y are the variables.

The standard form or general form of a linear equation is a x plus b y equals c

Linear Equation Graph

The graph of a linear equation in one variable x forms a vertical line that is parallel to the y-axis and vice-versa, whereas, the graph of a linear equation in two variables x and y forms a straight line. Let us graph a linear equation in two variables with the help of the following example.

Example: Plot a graph for a linear equation in two variables, x - 2y = 2.

Let us plot the linear equation graph using the following steps.

Step 1: The given linear equation is x - 2y = 2.
Step 2: Convert the equation in the form of y = mx + b. This will give: y = x/2 - 1.
Step 3: Now, we can replace the value of x for different numbers and get the resulting value of y to create the coordinates.
Step 4: When we put x = 0 in the equation, we get y = 0/2 - 1, i.e. y = -1. Similarly, if we substitute the value of x as 2 in the equation, y = x/2 - 1, we get y = 0.
Step 5: If we substitute the value of x as 4, we get y = 1. The value of x = -2 gives the value of y = -2. Now, these pairs of values of (x, y) satisfy the given linear equation y = x/2 - 1. Therefore, we list the coordinates as shown in the following table.
Step 6: Finally, we plot these points (4,1), (2,0), (0,-1) and (-2, -2) on a graph and join the points to get a straight line. This is how a linear equation is represented on a graph.

Graphical representation of Linear equations

Linear Equations in One Variable

A linear equation in one variable is an equation in which there is only one variable present. It is of the form Ax + B = 0, where A and B are any two real numbers and x is an unknown variable that has only one solution. It is the easiest way to represent a mathematical statement. This equation has a degree that is always equal to 1. A linear equation in one variable can be solved very easily. The variables are separated and brought to one side of the equation and the constants are combined and brought to the other side of the equation, to get the value of the unknown variable.

Example: Solve the linear equation in one variable: 3x + 6 = 18.

In order to solve the given equation, we bring the numbers on the right-hand side of the equation and we keep the variable on the left-hand side. This means, 3x = 18 - 6. Then, as we solve for x , we get, 3x = 12. Finally, the value of x = 12/3 = 4.

☛Also check:

Linear Equation In One Variable Questions
Linear Equations Word Problems

Linear Equations in Two Variables

A linear equation in two variables is of the form Ax + By + C = 0, in which A, B, C are real numbers and x and y are the two variables, each with a degree of 1. If we consider two such linear equations, they are called simultaneous linear equations. For example, 6x + 2y + 9 = 0 is a linear equation in two variables. There are various ways of solving linear equations in two variables like the graphical method , the substitution method , the cross multiplication method , the elimination method , and the determinant method.

☛Also check: Linear Equations In Two Variables Worksheets

How to Solve Linear Equations?

An equation is like a weighing balance with equal weights on both sides. If we add or subtract the same number from both sides of an equation, it still holds true. Similarly, if we multiply or divide the same number on both sides of an equation, it is correct. We bring the variables to one side of the equation and the constant to the other side and then find the value of the unknown variable. This is the way to solve a linear equation with one variable. Let us understand this with the help of an example.

Example: Solve the equation, 3x - 2 = 4.

We perform mathematical operations on the Left-hand side (LHS) and the right-hand side (RHS) so that the balance is not disturbed. So, let us add 2 on both sides to reduce the LHS to 3x. This will not disturb the balance. The new LHS is 3x - 2 + 2 = 3x and the new RHS is 4 + 2 = 6. Now, let us divide both sides by 3 to reduce the LHS to x. Thus, we have x = 2 . This is one of the ways of solving linear equations in one variable.

Tips on Linear Equations:

The value of the variable that makes a linear equation true is called the solution or root of the linear equation.
The solution of a linear equation is unaffected if the same number is added, subtracted, multiplied, or divided into both sides of the equation.
The graph of a linear equation in one or two variables always forms a straight line.

☛ Related Articles:

Introduction to Graphing
Linear Polynomial
Solving Linear Equations Calculator

Linear Equation Examples

Example 1: The sum of two numbers is 44. If one number is 10 more than the other, find the numbers by framing a linear equation.

Hint: This problem can be solved by writing linear equation in one variable.

Let the number be x, so the other number is x + 10. We know that the sum of both numbers is 44. Therefore, the linear equation can be framed as, x + x + 10 = 44. This results in, 2x + 10 = 44. Now, let us solve the equation by isolating the variable on one side and by bringing the constants on the other side. This means 2x = 44 - 10. By simplifying RHS, we get, 2x = 34, so the value of x is 17. This means, one number is 17 and the other number is 17 + 10 = 27.

Answer: Therefore, the two numbers are 17 and 27.

Example 2: Six times of a number is equal to 48. Find the linear equation that corresponds to the situation and find the unknown number.

Solution: Let the unknown number be x. Six times of this number is equal to 48. This gives the linear equation 6x = 48. So, this linear equation can be solved to find the value of x which is the unknown number. 6x = 48 means x = 48/6 = 8.

Answer: Therefore, the unknown number is 8.

Example 3: Calculate the linear equation for x: 5x - 95 = 75.

Solution: The given equation is 5x - 95 = 75.

⇒ 5x = 75 + 95

Answer: Therefore, the value of x is 34.

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FAQs on Linear Equation

What is a linear equation explain with an example..

A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. When this equation is graphed, it always results in a straight line . This is the reason why it is termed as a 'linear equation'. There are linear equations in one variable, in two variables, in three variables, and so on. A few examples of linear equations are 5x + 6 = 1, 42x + 32y = 60, 7x = 84, etc.

What is the Formula for a Linear Equation?

The formula for a linear equation is the way in which a linear equation is expressed. It can be expressed in the standard form, the slope-intercept form or the point-slope form . Using the slope-intercept form, the linear equation can be found using y = mx + c and using the point-slope form, it can be found using y - y 1 = m(x-x 1 ), where m is the slope, c is the y-intercept, and (x 1 , y 1 ) is a point on the line.

Why is a Linear Equation Called Linear?

A linear equation is called linear because when we try to plot the graph of the given linear function , it results in a straight line.

How do you Solve Linear Equations?

We can solve a linear equation in one variable by moving the variables to one side of the equation, and the numeric part on the other side. For example, x - 1 = 5 - 2x can be solved by moving the numeric parts on the right-hand side of the equation, while keeping the variables on the left side. Hence, we get x + 2x = 5 + 1. Thus, 3x = 6. This gives x = 2.

Can Linear Equations have Fractions?

Yes, linear equations can have fractions only as long as the denominator in the fractional part is a constant value. The variables cannot be a part of the denominator of any fraction in a linear equation.

What are Linear Equations in One Variable?

How do You Convert a Linear Equation to Standard Form?

To convert a linear equation to standard form, you need to move all the variables to one side of the equation and the constants to the other side, and then rearrange the terms so that the variables are on the left side and the constant is on the right side.

What are Linear Equations in two Variables?

A linear equation in two variables is of the form Ax + By + C = 0, in which A and B are the coefficients , C is a constant term, and x and y are the two variables, each with a degree of 1. For example, 7x + 9y + 4 = 0 is a linear equation in two variables. If we consider two such linear equations, they are called simultaneous linear equations .

How are Quadratic Equations Different from Linear Equations?

Linear equations do not have any exponent other than 1 in any term. The general form of a linear equation is expressed as Ax + By + C = 0, where A, B, and C are any real numbers and x and y are the variables. Whereas, quadratic equations have at least one term containing a variable that is raised to the second power. The general form of a quadratic equation is expressed as ax 2 + bx + c = 0. Another difference between the two types of equations is that a linear equation forms a straight line whereas a quadratic equation forms a parabola on the graph.

How to Graph Linear Equations?

When we graph linear equations, it forms a straight line. In order to graph an equation of the form, Ax + By = C, we get two solutions that are corresponding to the x-intercepts and the y-intercepts . We convert the equation to the form, y = mx + b . Then, we replace the value of x with different numbers and get the value of y which creates a set of (x,y) coordinates. These coordinates can be plotted on the graph and then joined by a line.

How to Solve Linear Equations with Fractions?

Linear equations with fractions are solved in the same way as we solve the usual equations. We need to bring the variable on one side and the constants on the other side and solve for the variable. For example, let us solve the equation (2a/3) - 10 = 12.

Step 1: Here, we will bring the constants on the right-hand side, i.e., (2a/3) = 12 + 10.
Step 2: Now, we have (2a/3) = 22. This can be further written as, 2a = 22 × 3.
Step 3: Therefore, the value of a = 66/2 = 33.

Solving Equations Practice Questions

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Word Problems Linear Equations

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

$\textbf{1)}$ Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations $y= 5x+105,\,\,\,y=15x+5$ Show Answer 10 weeks ($155)

$\textbf{2)}$ mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations $m+s=50,\,\,\,m=s+10$ show answer mike collected 30 rocks, sarah collected 20 rocks., $\textbf{3)}$ in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations $b+g=50,\,\,\,3b=2g$ show answer 15 girls (10 boys), $\textbf{4)}$ kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations $c=10x+100,\,\,\,r=30x$ show answer 5 sandals ($150), $\textbf{5)}$ molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), $\textbf{6)}$ anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations $c+d=40,\,\,\,5d=4c$ show answer 20 cats (16 dogs), $\textbf{7)}$ a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations $o+a=15,\,\,\,1o+2a=25$ show answer 10 apples and 5 oranges, $\textbf{8)}$ the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations $r+g=36,\,\,\,7r=2g$ show answer 8 red marbles (28 green marbles), $\textbf{9)}$ a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost $c$ in terms of $h$ hours playing tennis. show equation the equation is $c=10h+100$, $\textbf{10)}$ emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations $e = 10x + 80,\,\,\,l = 6x + 120$ show answer 10 weeks ($180 each), $\textbf{11)}$ mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations $m + l = 200,\,\,\,m = l + 40$ show answer mark collected 120 stamps, lisa collected 80 stamps., $\textbf{12)}$ in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations $b + g = 40,\,\,\,5b = 3g$ show answer 15 boys (25 girls), $\textbf{13)}$ lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations $c=60,\,\,\,r=20x$ show answer 3 pieces, $\textbf{14)}$ tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations $b + n = 12,\,\,\,15b+3n=120$ show answer 5 notebooks (7 books), $\textbf{15)}$ emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations $r + g = 36,\,\,\,5g=4r$ show answer 16 guinea pigs (20 rabbits), $\textbf{16)}$ mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations $m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20$ show answer they spent $90 in total., $\textbf{17)}$ the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations $r + b = 50,\,\,\,3r=2b$ show answer 30 blue marbles (20 red marbles), $\textbf{18)}$ a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations $l + s = 5,\,\,\,12l+8s=52$ show answer they bought 3 large pizzas (2 small pizzas)., $\textbf{19)}$ the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations $a=l\times w,\,\,\,l=8,\,\,\,a=48$ show answer the width is 6 meters., $\textbf{20)}$ two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations $x+y=50,\,\,\,x=y+10$ show answer the numbers are 30 and 20., $\textbf{21)}$ a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations $j+t=8,\,\,\,40j+20t=260$ show answer 5 jeans and 3 t-shirts were sold., $\textbf{22)}$ the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., $\textbf{23)}$ a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost $c$ in terms of $m$ minutes. show equation the equation is c=30+0.10m, $\textbf{24)}$ a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations $a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6$ show answer the area is 24 square inches., $\textbf{25)}$ a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations $t+p=4,\,\,\,25t+45p=180$ show answer 0 shirts and 4 pants were sold., $\textbf{26)}$ a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations $a=l\times w,\,\,\,l=12,\,\,\,w=10$ show answer the area is 120 square feet., $\textbf{27)}$ the sum of two consecutive odd numbers is 56. what are the two numbers show equations $x+y=56,\,\,\,x=y+2$ show answer the numbers are 27 and 29., $\textbf{28)}$ a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations $a+c=10,\,\,\,15a+5c=110$ show answer 6 action figures and 4 toy cars were sold., $\textbf{29)}$ a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations $p+c=14,\,\,\,2p+c=25$ show answer 11 pies and 3 cookies were sold., $\textbf{for 30-33}$ two car rental companies charge the following values for x miles. car rental a: $y=3x+150 \,\,$ car rental b: $y=4x+100$, $\textbf{30)}$ which rental company has a higher initial fee show answer company a has a higher initial fee, $\textbf{31)}$ which rental company has a higher mileage fee show answer company b has a higher mileage fee, $\textbf{32)}$ for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., $\textbf{33)}$ what does the $3$ mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., $\textbf{34)}$ what does the $100$ mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., $\textbf{for 35-39}$ andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. $g=12-\frac{m}{18}$, $\textbf{35)}$ what does the $12$ in the equation represent show answer andy has $12$ gallons in his car when he starts his drive., $\textbf{36)}$ what does the $18$ in the equation represent show answer it takes $18$ miles to use up $1$ gallon of gas., $\textbf{37)}$ how many miles until he runs out of gas show answer the answer is $216$ miles, $\textbf{38)}$ how many gallons of gas does he have after 90 miles show answer the answer is $7$ gallons, $\textbf{39)}$ when he has $3$ gallons remaining, how far has he driven show answer the answer is $162$ miles, $\textbf{for 40-42}$ joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., $\textbf{40)}$ find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is $y=.1(x-5,000)$, $\textbf{41)}$ how much does joe need to sell to earn $10,000 in a month. show answer the answer is $$105,000$, $\textbf{42)}$ how much does joe earn if he sells $45,000 in a month show answer the answer is $$4,000$, see related pages, $\bullet\text{ word problems- linear equations}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- averages}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- consecutive integers}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- distance, rate and time}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- break even}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- ratios}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- age}$ $\,\,\,\,\,\,\,\,$, $\bullet\text{ word problems- mixtures and concentration}$ $\,\,\,\,\,\,\,\,$, linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

Linear Equation in One Variable Questions

Students will learn the concepts more effectively with the help of the linear equation in one variable questions and answers. Almost every class includes a discussion of linear equations.. The questions will be prepared in accordance with the NCERT norms. Both in daily life and in mathematics, linear equations are used. So, it’s important to understand the fundamentals of linear equations in one variable. The problems presented here will cover both the fundamentals and more difficult problems for students of all skill levels. Click here to read more about linear equations in one variable .

Go through the below problems to understand the linear equation in one variable and also solve the practice problems.

Linear Equation in One Variable Questions with Solutions

1. Convert the given statement into equation: An integer increased by 5 equals 20.

Given statement: An integer increased by 5 equals 20.

Assume that the unknown number is “x”.

Hence, x increased by 5 means x + 5.

Therefore, the equation x + 5 = 20

So, an integer increased by 5 equals 20 means x + 5 = 20, which is the required equation for the given statement.

2. Convert the given equation into a statement: 5x = 25.

The equation given is 5x = 25.

The statement equivalent to the equation 5x = 25 is “Five times the number x equals 25”.

3. Find the value of x for the equation 2x – 7 = 21. Also, verify the answer.

Given equation:2x – 7 = 21.

To solve the given equations, keep the variable x on one side and the constants on the other side.

Hence, the given equation becomes:

2x = 21 + 7

Therefore, the value of x is 14.

Verification:

Substitute x = 14 in the equation 2x – 7 = 21

Hence, 2(14) – 7 = 21

28 – 7 = 21

Hence, LHS = RHS.

4. Check whether x = 40 is the solution of the equation 5x/2 = 100.

To check whether x = 40 is the solution or root of the equation 5x/2 = 100, put the value x = 40 in the equation 5x/2 = 100

⇒ [5(40)]/2 = 100

⇒ 200/2 = 100

⇒ 100 = 100

Since, LHS = RHS, x = 40 is the solution of the equation 5x/2 = 100.

5. Find the number, if 15 is added to three times of the number results in 45. Also, justify your answer.

Given statement: 15 is added to three times of the number resulting in 45.

Let the unknown number be x.

Then according to the given statement, the equation formed is:

15 + 3x = 45

Now, keep the variable “x” on one side of the equation and constant on the other side of the equation.

⇒ 3x = 45 -15

Therefore, the required number is x = 10.

Substitute the value x =10 in the equation 15 + 3x = 45.

⇒ 15 + 3(10) = 45

⇒ 15 + 30 = 45

Therefore, LHS = RHS.

Hence, verified.

Also, read: Linear Equations in Two Variables.

6. Check whether the given statements are true or false.

If x = 5, then 5x – 5 = 20
If x = 7, then 4x – 4 = 20.

(a) Given equation: 5x – 5 = 20

= 5(5) – 5

= 25 – 5

Hence, if x = 5, then 5x – 5 = 20.

Therefore, the given equation is true.

(b) Given equation: 4x – 4 = 20

= 4(7) – 4

= 28 – 4

Hence, if x = 7, then 4x – 4 = 24.

Therefore, the given equation is false..

7. Determine the number, if the sum of two odd consecutive numbers is 56.

As we know that the difference between two odd consecutive numbers is 2.

Hence, let the small odd numbers be x and the successive odd numbers is x+2.

According to the given condition, we can write

⇒ x + x + 2 = 56

⇒ 2x + 2 = 56

⇒ 2x = 56 – 2

⇒ x = 27, which is the required odd number.

Hence, the other odd number is:

⇒ x + 2 = 27 + 2 = 29.

Therefore, the two odd numbers are 27 and 29.

8. Solve the given linear equation: 17 + 6p = 9

Given linear equation: 17 + 6p = 9

Now, keep the variable “p” on the left hand side and bring the constants on the right hand side.

Hence, we get

⇒ 6p = 9 – 17

Therefore, the value of p is -4/3.

Also, read: How to Solve Linear Equations?

9. Find the two numbers, if the numbers are in the ratio 5 : 3 and they differ by 18.

Let the unknown number be “p”.

Also, given that, the two numbers are in the ratio 5 : 3.

According to given conditions, we can write

⇒ 5p – 3p = 18

Therefore, the two numbers are:

5p = 5(9) = 45

3p = 3(9) = 27

Hence, the required two numbers are 45 and 27.

10. The ages of Rahul and Ramya are in the ratio 5: 7. After 4 years, the sum of their ages will be 56 years. Find their present ages.

Assume that the ages of Rahul and Ramya are 5p and 7p.

After 4 years, the ages of Rahul and Ramya will be 5p + 4 and 7p + 4.

According to the given condition, we get the following equation:

(5p + 4) + (7p + 4) = 56

⇒ 12p + 8 = 56

⇒ 12p = 56 – 8

⇒ p = 48/12

Therefore, Rahul’s present age = 5(4) = 20

Ramya’s present age = 7(4) = 28.

Hence, the present age of Rahul and Ramya are 20 and 28, respectively.

Explore More Articles

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Practice Questions

Write the equation for the given statement: 3 decreased from 4x gives 17.
Convert the given equation into a statement: 9x – 9 = 81.
Determine the value of x for the given equation: 25x + 100 = 350.

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1.20: Word Problems for Linear Equations

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Word problems are important applications of linear equations. We start with examples of translating an English sentence or phrase into an algebraic expression.

Example 18.1

Translate the phrase into an algebraic expression:

a) Twice a variable is added to 4

Solution: We call the variable $x .$ Twice the variable is $2 x .$ Adding $2 x$ to 4 gives:

\[4 + 2x\nonumber\]

b) Three times a number is subtracted from 7.

Solution: Three times a number is $3 x .$ We need to subtract $3 x$ from 7. This means:\

\[7-3 x\nonumber\]

c) 8 less than a number.

Solution: The number is denoted by $x .8$ less than $x$ mean, that we need to subtract 8 from it. We get:

\[x-8\nonumber\]

For example, 8 less than 10 is $10-8=2$.

d) Subtract $5 p^{2}-7 p+2$ from $3 p^{2}+4 p$ and simplify.

Solution: We need to calculate $3 p^{2}+4 p$ minus $5 p^{2}-7 p+2:$

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)\nonumber\]

Simplifying this expression gives:

\[\left(3 p^{2}+4 p\right)-\left(5 p^{2}-7 p+2\right)=3 p^{2}+4 p-5 p^{2}+7 p-2 =-2 p^{2}+11 p-2\nonumber\]

e) The amount of money given by $x$ dimes and $y$ quarters.

Solution: Each dime is worth 10 cents, so that this gives a total of $10 x$ cents. Each quarter is worth 25 cents, so that this gives a total of $25 y$ cents. Adding the two amounts gives a total of

\[10 x+25 y \text{ cents or } .10x + .25y \text{ dollars}\nonumber\]

Now we deal with word problems that directly describe an equation involving one variable, which we can then solve.

Example 18.2

Solve the following word problems:

a) Five times an unknown number is equal to 60. Find the number.

Solution: We translate the problem to algebra:

\[5x = 60\nonumber\]

We solve this for $x$ :

\[x=\frac{60}{5}=12\nonumber\]

b) If 5 is subtracted from twice an unknown number, the difference is $13 .$ Find the number.

Solution: Translating the problem into an algebraic equation gives:

\[2x − 5 = 13\nonumber\]

We solve this for $x$. First, add 5 to both sides.

\[2x = 13 + 5, \text{ so that } 2x = 18\nonumber\]

Dividing by 2 gives $x=\frac{18}{2}=9$.

c) A number subtracted from 9 is equal to 2 times the number. Find the number.

Solution: We translate the problem to algebra.

\[9 − x = 2x\nonumber\]

We solve this as follows. First, add $x$ :

\[9 = 2x + x \text{ so that } 9 = 3x\nonumber\]

Then the answer is $x=\frac{9}{3}=3$

d) Multiply an unknown number by five is equal to adding twelve to the unknown number. Find the number.

Solution: We have the equation:

\[5x = x + 12.\nonumber\]

Subtracting $x$ gives

\[4x = 12.\nonumber\]

Dividing both sides by 4 gives the answer: $x=3$.

e) Adding nine to a number gives the same result as subtracting seven from three times the number. Find the number.

Solution: Adding 9 to a number is written as $x+9,$ while subtracting 7 from three times the number is written as $3 x-7$. We therefore get the equation:

\[x + 9 = 3x − 7.\nonumber\]

We solve for $x$ by adding 7 on both sides of the equation:

\[x + 16 = 3x.\nonumber\]

Then we subtract $x:$

\[16 = 2x.\nonumber\]

After dividing by $2,$ we obtain the answer $x=8$

The following word problems consider real world applications. They require to model a given situation in the form of an equation.

Example 18.3

a) Due to inflation, the price of a loaf of bread has increased by $5 \%$. How much does the loaf of bread cost now, when its price was $\$ 2.40$ last year?

Solution: We calculate the price increase as $5 \% \cdot \$ 2.40 .$ We have

\[5 \% \cdot 2.40=0.05 \cdot 2.40=0.1200=0.12\nonumber\]

We must add the price increase to the old price.

\[2.40+0.12=2.52\nonumber\]

The new price is therefore $\$ 2.52$.

b) To complete a job, three workers get paid at a rate of $\$ 12$ per hour. If the total pay for the job was $\$ 180,$ then how many hours did the three workers spend on the job?

Solution: We denote the number of hours by $x$. Then the total price is calculated as the price per hour $(\$ 12)$ times the number of workers times the number of hours $(3) .$ We obtain the equation

\[12 \cdot 3 \cdot x=180\nonumber\]

Simplifying this yields

\[36 x=180\nonumber\]

Dividing by 36 gives

\[x=\frac{180}{36}=5\nonumber\]

Therefore, the three workers needed 5 hours for the job.

c) A farmer cuts a 300 foot fence into two pieces of different sizes. The longer piece should be four times as long as the shorter piece. How long are the two pieces?

\[x+4 x=300\nonumber\]

Combining the like terms on the left, we get

\[5 x=300\nonumber\]

Dividing by 5, we obtain that

\[x=\frac{300}{5}=60\nonumber\]

Therefore, the shorter piece has a length of 60 feet, while the longer piece has four times this length, that is $4 \times 60$ feet $=240$ feet.

d) If 4 blocks weigh 28 ounces, how many blocks weigh 70 ounces?

Solution: We denote the weight of a block by $x .$ If 4 blocks weigh $28,$ then a block weighs $x=\frac{28}{4}=7$

How many blocks weigh $70 ?$ Well, we only need to find $\frac{70}{7}=10 .$ So, the answer is $10 .$

Note You can solve this problem by setting up and solving the fractional equation $\frac{28}{4}=\frac{70}{x}$. Solving such equations is addressed in chapter 24.

e) If a rectangle has a length that is three more than twice the width and the perimeter is 20 in, what are the dimensions of the rectangle?

Solution: We denote the width by $x$. Then the length is $2 x+3$. The perimeter is 20 in on one hand and $2($length$)+2($width$)$ on the other. So we have

\[20=2 x+2(2 x+3)\nonumber\]

Distributing and collecting like terms give

\[20=6 x+6\nonumber\]

Subtracting 6 from both sides of the equation and then dividing both sides of the resulting equation by 6 gives:

\[20-6=6 x \Longrightarrow 14=6 x \Longrightarrow x=\frac{14}{6} \text { in }=\frac{7}{3} \text { in }=2 \frac{1}{3} \text { in. }\nonumber\]

f) If a circle has circumference 4in, what is its radius?

Solution: We know that $C=2 \pi r$ where $C$ is the circumference and $r$ is the radius. So in this case

\[4=2 \pi r\nonumber\]

Dividing both sides by $2 \pi$ gives

\[r=\frac{4}{2 \pi}=\frac{2}{\pi} \text { in } \approx 0.63 \mathrm{in}\nonumber\]

g) The perimeter of an equilateral triangle is 60 meters. How long is each side?

Solution: Let $x$ equal the side of the triangle. Then the perimeter is, on the one hand, $60,$ and on other hand $3 x .$ So $3 x=60$ and dividing both sides of the equation by 3 gives $x=20$ meters.

h) If a gardener has $\$ 600$ to spend on a fence which costs $\$ 10$ per linear foot and the area to be fenced in is rectangular and should be twice as long as it is wide, what are the dimensions of the largest fenced in area?

Solution: The perimeter of a rectangle is $P=2 L+2 W$. Let $x$ be the width of the rectangle. Then the length is $2 x .$ The perimeter is $P=2(2 x)+2 x=6 x$. The largest perimeter is $\$ 600 /(\$ 10 / f t)=60$ ft. So $60=6 x$ and dividing both sides by 6 gives $x=60 / 6=10$. So the dimensions are 10 feet by 20 feet.

i) A trapezoid has an area of 20.2 square inches with one base measuring 3.2 in and the height of 4 in. Find the length of the other base.

Solution: Let $b$ be the length of the unknown base. The area of the trapezoid is on the one hand 20.2 square inches. On the other hand it is $\frac{1}{2}(3.2+b) \cdot 4=$ $6.4+2 b .$ So

\[20.2=6.4+2 b\nonumber\]

Multiplying both sides by 10 gives

\[202=64+20 b\nonumber\]

Subtracting 64 from both sides gives

\[b=\frac{138}{20}=\frac{69}{10}=6.9 \text { in }\nonumber\]

and dividing by 20 gives

Exit Problem

Write an equation and solve: A car uses 12 gallons of gas to travel 100 miles. How many gallons would be needed to travel 450 miles?

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SAT Mathematics : Solving Linear Equations

Study concepts, example questions & explanations for sat mathematics, all sat mathematics resources, example questions, example question #1 : solving linear equations.