 # Solving Systems of Linear Equations By Elimination

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Solving Systems of Linear Equations By Elimination

What is Elimination? To eliminate means to get rid of or remove.
You solve equations by eliminating one of the variables (x or y) using addition or subtraction.

Example 1 Solve the following system of linear
equations by elimination. 2x – 3y = 15 5x + 3y = 27 (1) (2) Add equation (1) to equation (2) 7x + 0y = 42 7x = 42  By eliminating y, we can now solve for x x = 6

Example 1 Substitute x= 6 into equation (1) to solve for y
Check your solution x = 6 and y = -1 in equation (2) 2x – 3y = 15 5x + 3y = 27 2(6) – 3y = 15 5(6) + 3(-1) = 27 30 – 3 = 27 12 – 3y = 15 27 = 27 – 3y = 15 – 12 LS = RS – 3y = 3 y = -1 Therefore, the solution set = (6,-1)

You Try 5x – 6y = -32 3x + 6y = 48

One More 7x + 2y = −19 −x + 2y = 21

If you have noticed in the last few examples that to eliminate a variable, it’s coefficients must have a sum or difference of zero. Sometimes you may need to multiply one or both of the equations by a nonzero number first so that you can then add or subtract the equations to eliminate one of the variables. 2x + 5y = 17 7x + 2y = 10 2x + 5y = -22 6x – 5y = x + y = x + 3y = 22 We can add these two equations together to eliminate the x variable. We can add these two equations together to eliminate the y variable. What are we going to do with these equations, can’t eliminate a variable the way they are written?

Multiplying One Equation
Solve by Elimination 2x + 5y = -22 10x + 3y = 22 2x + 5y = (2x + 5y = -22) x y = -110 10x + 3y = x + 3y = (10x + 3y = 22) y = -132 y = -6

Step y = -6 Solve for the eliminated variable using either of the original equations x + 5y = -22 Choose the first equation. 2x + 5(-6) = -22 Substitute -6 for y. 2x – 30 = Solve for x x = x = 4 The solution is (4, -6).

Solve by elimination. -2x + 5y = -32 7x – 5y = x – 3y = 61 2x + y = x – 10y = -25 4x + 40y = 20

5x + 4y = -28 3x + 10y = -13

Multiplying Both Equations
To eliminate a variable, you may need to multiply both equations in a system by a nonzero number. Multiply each equation by values such that when you write equivalent equations, you can then add or subtract to eliminate a variable. 4x + 2y = 14 7x + 3y = -8 In these two equations you cannot use graphing or substitution very easily. However ever if we multiply the first equation by 3 and the second by 2, we can eliminate the y variable. Find the least common multiple LCM of the coefficients of one variable, since working with smaller numbers tends to reduce the likelihood of errors. 4 x 7 = 28 2 x 3 = 6

4x + 2y = 14 3(4x + 2y = 14) x + 6y = 42 7x – 3y = (7x – 3y = -8) x – 6y = x = x = x = 1 Solve for the eliminated variable y using either of the original equations. 4x + 2y = 14 4(1) + 2y = y = y = y = 5 The solution is (1, 5).

Solve by your method of choice: 1) 2x + 5y = 17 2) 7x + 2y = 10
Closure Solve by your method of choice: 1) 2x + 5y = ) 7x + 2y = 10 6x + 5y = x + y = -16 3) 2x – 3y = ) 24x + 2y = 52 2x + y = x – 3y = -36 5) y = 2x 6) 9x + 5y = 34 y = x – x – 2y = -2 1. (1, 3) 2. (2, -2) 3. (5, -17) 4. (1, 14) 5. (-1, -2) 6. (1, 5)