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Linear Equations in Two Variables

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Presentation on theme: "Linear Equations in Two Variables"— Presentation transcript:

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LINEAR EQUATIONS IN TWO VARIABLES - PowerPoint PPT Presentation

LINEAR EQUATIONS IN TWO VARIABLES

Linear equations in two variables system of equations or simultaneous equations a pair of linear equations in two variables is said to form a system of ... – powerpoint ppt presentation.

• Introduction
• A simple linear equation is an equality between two algebraic expressions involving an unknown value called the variable. In a linear equation the exponent of the variable is always equal to 1. The two sides of an equation are called Right Hand Side (RHS) and Left-Hand Side (LHS). They are written on either side of equal sign. LHS RHS EX 4x3 15 2x5y 0 -2x3y 6
• System of equations or simultaneous equations
• A pair of linear equations in two variables is said to form a system of simultaneous linear equations.
• For Example, 2x 3y 4 0
• Form a system of two linear equations in variables x and y.
• GENERAL FORM
• The general form of a linear equation in two variables xand y is
• ax by c 0 , a / 0, b/0, where
• a, b and c being real numbers.
• A solution of such an equation is a pair of values, one for x and the other for y, which makes two sides of the equation equal.
• Every linear equation in two variables has infinitely many solutions which can be represented on a certain line.
• Firstly we have to know that whether the equations can be solved or not. For this we have the rules shown below -
• 1. a1 / b1 (UNIQUE SOLUTION)
• 2. a1 b1 c1 (INFINITE SOLUTIONS)
• 3. a1 b1 / c1 (NO SOLUTION)
• The most commonly used algebraic methods of solving simultaneous linear equations in two variables are
• Method of substitution
• Method of elimination
• Method of Cross- multiplication
• Obtain the two equations. Let the equations be
• a1x b1y c1 0 ----------- (i)
• a2x b2y c2 0 ----------- (ii)
• Choose either of the two equations, say (i) and find the value of one variable , say y in terms of x
• Substitute the value of y, obtained in the previous step in equation (ii) to get an equation in x
• Solve the equation obtained in the previous step to get the value of x.
• Substitute the value of x and get the value of y.
• Let us take an example
• x 2y -1 ------------------ (i)
• 2x 3y 12 -----------------(ii)
• x -2y -1 ------- (iii)
• Substituting the value of x in equation (ii), we get
• 2 ( -2y 1) 3y 12
• - 4y 2 3y 12
• - 7y 14 , y -2 ,
• Putting the value of y in eq (iii), we get
• x - 2 x (-2) 1
• Hence the solution of the equation is
• In this method, we eliminate one of the two variables to obtain an equation in one variable which can easily be solved. Putting the value of this variable in any of the given equations, the value of the other variable can be obtained.
• We eliminate one variable first , to get a linear equation in one variable.
• Step 1. first multiply both the equation by some suitable non-zero constants to make the coefficients of one variable numerically equal.
• Step 2. then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to step
• Step 3. solve equation in one variable so obtained to get its value.
• For example we want to solve,
• Let 3x 2y 11 --------- (i)
• 2x 3y 4 ---------(ii)
• Multiply 3 in equation (i) and 2 in equation (ii) and subtracting eqn. iv from iii, we get
• 9x 6y 33 ------ (iii)
• 4x 6y 8 ------- (iv)
• putting the value of y in equation (ii) we get,
• Hence, x 5 and y -2
• This is a method very useful for solving the linear equation in two variables
• Let us consider two equations-a1x b1y c1 0
• a2x b2y c2 0
• b1 c1 a1 b1
• b2 c2 a2 b2
• b1c2 b2c1 c1a2 c2a1 a1b2 a2b1
• By this way the equations are solved and the values are obtained.
• We have to put the values of the known and get the values of the unknown.
• We can write this as given below also
• x b1c2 - b2c1
• a1b2 - a2b1
• y c1a2 - c2a1
• The equations which cannot be solved simply are converted to the reduced forms and then solved such as -
• 2/x 3/y 13 ? 2(1/x) 3(1/y) 13
• 5/x 4/y -2 ? 5(1/x) 4(1/y) -2
• Let (1/x) p (1/y) q, then the equations - 2p 3q 13 5p 4q -2 can be solved by any of the three methods mentioned above.
• FOMATIVE ASSESSMENT(MCQ)
• 1. Which of the following is the solution of the pair of linear equations 3x 2y 0, 5y x 0
• (a) (5, 1) (b) (2, 3) (c) (1, 5) (d) (0, 0)
• 2. One of the common solution of ax by c and y-axis is _____
• (a) (0, c/b) (b) (0,b/c ) (c) , 0 , (c/ b ) (d) (0, c/ b)
• 3. If the value of x in the equation 2x 8y 12 is 2 then the corresponding value of y will be
• (a) 1 (b) 1 (c) 0 (d) 2
• To verify graphically
• That the pair of linear equations xy-50 ,2x2y-60 in which a1/a2b1/b2 / c1/c2 gives a pair of parallel straight lines.
• x y 5 ..(1)
• SOLUTION FOR XY5
• SOLUTION FOR 2X2Y6
• After plotting the points on graph we get

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Systems of Linear Equations in Two Variables

Mar 13, 2019

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Systems of Linear Equations in Two Variables. Systems of Linear Equations and Their Solutions.

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• back substitute
• ordered pair
• step 1 solve
• step 5 back substitute

Presentation Transcript

Systems of Linear Equations and Their Solutions We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations, such as those listed above, are called a system of linear equations. A solution to a system of linear equations is an ordered pair that satisfies all equations in the system. For example, (3, 4) satisfies the system x + y = 7 (3 + 4 is, indeed, 7.) x – y = -1 (3 – 4 is indeed, -1.) Thus, (3, 4) satisfies both equations and is a solution of the system. The solution can be described by saying that x = 3 and y = 4. The solution can also be described using set notation. The solution set to the system is {(3, 4)} - that is, the set consisting of the ordered pair (3, 4).

Solution Because 4 is the x-coordinate and -1 is the y-coordinate of (4, -1), we replace x by 4 and y by -1. x + 2y = 2 x – 2y = 6 4 + 2(-1) = 2 4 – 2(-1) = 6 4 + (-2) = 2 4 – (-2) = 6 2 = 2 true 4 + 2 = 6 6 = 6 true The pair (4, -1) satisfies both equations: It makes each equation true. Thus, the pair is a solution of the system. The solution set to the system is {(4, -1)}. ? ? ? ? ? Text Example Determine whether (4, -1) is a solution of the system x + 2y = 2 x – 2y = 6.

Solving Linear Systems by Substitution • Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.) • Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable. • Solve the equation obtained in step 2. • Back-substitute the value found in step 3 into the equation from step 1. Simplify and find the value of the remaining variable. • Check the proposed solution in both of the system's given equations.

Step 2Substitute the expression from step 1 into the other equation. We substitute 2y - 3 for x in the first equation. x = 2y – 3 5 x – 4y = 9 Text Example Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution Step 1Solve either of the equations for one variable in terms of the other. We begin by isolating one of the variables in either of the equations. By solving for x in the second equation, which has a coefficient of 1, we can avoid fractions. x - 2y = -3 This is the second equation in the given system. x = 2y - 3 Solve for x by adding 2y to both sides.

Text Example cont. Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution This gives us an equation in one variable, namely 5(2y - 3) - 4y = 9. The variable x has been eliminated. Step 3Solve the resulting equation containing one variable. 5(2y – 3) – 4y = 9 This is the equation containing one variable. 10y – 15 – 4y = 9 Apply the distributive property. 6y – 15 = 9 Combine like terms. 6y = 24 Add 15 to both sides. y = 4 Divide both sides by 6.

Text Example cont. Solve by the substitution method: 5x – 4y = 9 x – 2y = -3. Solution Step 4Back-substitute the obtained value into the equation from step 1. Now that we have the y-coordinate of the solution, we back-substitute 4 for y in the equation x = 2y – 3. x = 2y – 3 Use the equation obtained in step 1. x = 2 (4) – 3 Substitute 4 for y. x = 8 – 3 Multiply. x = 5 Subtract. With x = 5 and y = 4, the proposed solution is (5, 4). Step 5Check. Take a moment to show that (5, 4) satisfies both given equations. The solution set is {(5, 4)}.

Solving Linear Systems by Addition • If necessary, rewrite both equations in the form Ax + By = C. • If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0. • Add the equations in step 2. The sum is an equation in one variable. • Solve the equation from step 3. • Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable. • Check the solution in both of the original equations.

Solution Step 1Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain 2x - 7y = -17 3x + 5y = 17 Text Example Solve by the addition method: 2x = 7y - 17 5y = 17 - 3x. Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0. We can eliminate x or y. Let's eliminate x by multiplying the first equation by 3 and the second equation by -2.

Multiply by 3. 2x – 7y = -17 3•2x – 3•7y = 3(-17) 6x – 21y = -51 Multiply by -2. 3x + 5y = 17 -2•3x + (-2)5y = -2(17) -6x – 10y = -34 6x – 21y = -51 -6x – 10y = -34 -31y = -85 -31y = -85 -31 -31 y = 85/31 Text Example cont. Solution Steps 3 and 4 Add the equations and solve for the remaining variable. Add: Divide both sides by -31. Simplify. Step 5Back-substitute and find the value for the other variable. Back-substitution of 85/31 for y into either of the given equations results in cumbersome arithmetic. Instead, let's use the addition method on the given system in the form Ax + By = C to find the value for x. Thus, we eliminate y by multiplying the first equation by 5 and the second equation by 7.

Multiply by 5. 2x – 7y = -17 5•2x – 5•7y = 5(-17) 10x – 35y = -85 Multiply by 7. 3x + 5y = 17 7•3x + 7•5y = 7(17) 21x + 35y = 119 31x = 34 x = 34/31 Text Example cont. Solution Add: Step 6Check. For this system, a calculator is helpful in showing the solution (34/31, 85/31) satisfies both equations. Consequently, the solution set is {(34/31, 85/31)}.

y y y x x x Exactly one solution No Solution (parallel lines) Infinitely many solutions (lines coincide) The Number of Solutions to a System of Two Linear Equations The number of solutions to a system of two linear equations in two variables is given by one of the following. Number of SolutionsWhat This Means Graphically Exactly one ordered-pair solution The two lines intersect at one point. No solution The two lines are parallel. Infinitely many solutions The two lines are identical.

Solve the system 2x + 3y = 4 -4x - 6y = -1 Solution: 2 (2x + 3y = 4) multiply the first equation by 2-4x - 6y = -1 4x + 6y = 8-4x - 6y = -1 0 = 7 Add the two equationsNo solution Example

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