# Solving Linear Systems By Elimination. Solving Linear Systems There are three methods for solving a system of equations: By Graphing them and looking.

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Solving Linear Systems By Elimination

Solving Linear Systems There are three methods for solving a system of equations: By Graphing them and looking at the results By solving them algebraically using Elimination By solving them algebraically using Substituion

Elimination ● The key to solving a system by elimination is getting rid of one variable. ● Let’s review the Additive Inverse Property. ● What is the Additive Inverse of: 3x? -5y? 8p? -3x 5y -8p ● What happens if we add two additive inverses? We get zero. The terms cancel, therefore Eliminating them.

The Elimination Method ● We will try to eliminate one variable by adding, subtracting, or multiplying the variable(s) until the two terms are additive inverses. ● We will then add the two equations, giving us one equation with one variable. ● Solve for that variable. ● Then insert the value into one of the original equations to find the other variable.

Solving a System by Elimination ● Solve the system: x + y = 7 x - y = 5 ● Notice that the y terms in both equations are additive inverses. So if we add the equations the y terms will cancel. ● So let’s add & solve:x + y = 7 + x - y = 5 2x + 0 = 12 2x = 12 x = 6 ● Insert the value of x to find y: 6 + y = 7 so y = 1 The solution is (6, 1).

Solving a System by Elimination ● Solve the system: 2x + 3y = 23 x - 3y = -11 ● Notice that the y terms in both equations are additive inverses. So if we add the equations the y terms will cancel. ● So let’s add & solve:2x + 3y = 23 + x - 3y = -11 3x + 0 = 12 3x = 12 x = 4 ● Insert the value of x to find y: 2(4) + 3y = 23 so y = 5 The solution is (4, 5).

Another Elimination Example ● Solve the system: 3s - 2t = 10 4s + t = 6 ● What is the problem? There is NO additive inverse! ● We could multiply the second equation by 2 and the t terms would be inverses. Let’s multiply the second equation by 2 to eliminate t. 3s - 2t = 10 3s – 2t = 10 2(4s + t = 6) 8s + 2t = 12 ● Add and solve: 11s + 0 = 22 11s = 22 s = 2 ● Insert the value of s to find the value of t: 3(2) - 2t = 10 so t = -2 ● The solution is (2, -2).

Some to try: 1. -4x + y = -12 4x + 2y = 6 2.5x + 2y = 12 -6x -2y = -14 3.5x + 4y = 12 7x - 6y = 40 4.5m + 2n = -8 4m +3n = 2

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