# 3-2: Solving Linear Systems

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3-2: Solving Linear Systems

Solving Linear Systems
There are two methods of solving a system of equations algebraically: Elimination Substitution

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 y -8p -q What happens if we add two additive inverses? We get zero. The terms cancel. 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.

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).

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).

Elimination Solve the system by elimination: 1. -4x + y = -12
4. 5m + 2n = -8 4m +3n = 2

Substitution To solve a system of equations by substitution…
1. Solve one equation for one of the variables. 2. Substitute the value of the variable into the other equation. Simplify and solve the equation. 4. Substitute back into either equation to find the value of the other variable.

Substitution Solve the system: x - 2y = -5 y = x + 2
Notice: One equation is already solved for one variable. Substitute (x + 2) for y in the first equation. x - 2y = -5 x - 2(x + 2) = -5 We now have one equation with one variable. Simplify and solve. x - 2x – 4 = -5 -x - 4 = -5 -x = -1 x = 1 Substitute 1 for x in either equation to find y. y = x y = so y = 3 The solution is (1, 3).

Substitution Let’s check the solution. The answer (1, 3) must check
in both equations. x - 2y = y = x + 2 1 - 2(3) = = 1 + 2 -5 = -5 3 = 3 

Substitution Solve the system: 2p + 3q = 2 p - 3q = -17
Notice that neither equation is solved for a variable. Since p in the second equation does not have a coefficient, it will be easier to solve. p = 3q – 17 Substitute the value of p into the first equation, and solve. 2p + 3q = 2 2(3q – 17) + 3q = 2 6q – q = 2 9q – 34 = 2 9q = 36 q = 4

Substitution 2p + 3q = 2 p – 3q = -17
Substitute the value of q into the second equation to find p. p = 3q – 17 p = 3(4) – 17 p = -5 The solution is (-5, 4). (List p first since it comes first alphabetically.) Let’s check the solution: 2p + 3q = 2 p – 3q = -17 2(-5) +3(4) = (4) = -17 = = -17 2 = 2 = -17 

Substitution Solve the systems by substitution: 1. x = 4 2x - 3y = -19