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10.3 Solving Linear Systems

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Presentation on theme: "10.3 Solving Linear Systems"— Presentation transcript:

1 10.3 Solving Linear Systems
Elimination Method #3

2 All 3 Methods Graphing Substitution Elimination Solutions
Using addition/subtraction Using multiplication Solutions One solution No solution Infinitely many solutions

3 Elimination 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

4 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? q? -3x 5y p -q What happens if we add two additive inverses? We get zero. The terms cancel.

5 Elimination Solve the system: m + n = 6 m - n = 5
Notice that the n terms in both equations are additive inverses. So if we add the equations the n terms will cancel. So let’s add & solve: m + n = m - n = 5 2m + 0 = 11 2m = 11 m = 11/2 or 5.5 Insert the value of m to find n: n = 6 n = .5 The solution is (5.5, .5).

6 Elimination Solve the system: 3s - 2t = 10 4s + t = 6
We could multiply the second equation by 2 and the t terms would be inverses OR We could multiply the first equation by 4 and the second equation by -3 to make the s terms inverses. Let’s multiply the second equation by 2 to eliminate t. (It’s easier.) 3s - 2t = s – 2t = 10 2(4s + t = 6) s + 2t = 12 Add and solve: s + 0t = 22 11s = 22 s = 2 Insert the value of s to find the value of t (2) - 2t = t = -2 The solution is (2, -2).

7 Elimination 4m +3n = 2 Solve the system by elimination:
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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