Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 3 - Linear Systems - 3-2 Solving Systems Algebraically - Practice and Problem-Solving Exercises - Page 146: 26

Answer

The solution to this system of equations is $w = -2$ and $y = -4$.

Work Step by Step

We see that in the two equations, the $y$ term is the exactly the same except they have opposite signs. If we add these two equations together, we can eliminate the variable $y$ and just deal with one variable instead of two: $(2w+5y)+(3w-5y)=-24+14\\ 5w=-10$ Divide each side by $5$ to solve for $w$: $w = -2$ Now that we have the value for $w$, we can plug it into one of the equations to solve for $y$. Let us plug the value for $w$ into the second equation: $3(-2) - 5y = 14$ $-6 - 5y = 14$ Now, we add $6$ to both sides of the equation to isolate constants to the right side of the equation: $-5y = 20$ Divide both sides by $-5$ to solve for $y$: $y = -4$ The solution to this system of equations is $w = -2$ and $y = -4$.
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