Solve Systems of Linear Equations Substitution

Solve Systems Substitution strategies Redefine one variable in terms of the other. Substitute this new definition into the other equation. Solve for the single variable Replace the value for the known variable and solve for the other. Check both solutions

Which variable to redefine Choose a variable with a coefficient of one. Isolate this variable and use the variables value to substitute.

Types of solutions (again—three types) We will have one solution. No solutions (inconsistent). Infinite solutions (the two equations are the same).

2 2𝑦−5 +4𝑦=−6 (now distribute) 4𝑦−10+4𝑦=−6 (combine like terms) 8𝑦=4 An example– (I do) Use the definition 𝑥=−2𝑦−5 and substitute this into the other equation. 2 2𝑦−5 +4𝑦=−6 (now distribute) 4𝑦−10+4𝑦=−6 (combine like terms) 8𝑦=4 𝑦= 1 2

We know 𝑦= 1 2 . Now we substitute this in either equation and solve for x. Choose the easiest to solve before substituting! 𝑥=2 1 2 −5 𝑥=−4

Check the solutions in both equations 2 −4 +4 1 2 =−6 (checks) −4=2 1 2 −5 (checks)

One more (we do) solve Use the first equation to redefine y 𝑦=𝑥+1 (now substitute into equation 2 2𝑥+(𝑥+1)=−2 (combine like terms) 3𝑥=−3 𝑥=−1

In either equation, replace x with -1 (we’ll use the second) 2 −1 +𝑦=−2 𝑦=0

Check both equations with the solutions. (x=-1 & y = 0) −−1+0=1 (checks) 2 −1 +0=−2 (checks)