Algebra 2 Common Core

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

Chapter 3 - Linear Systems - Chapter Review - Page 184: 12

Answer

The solution to this system of equations is $(-2, 3)$.

Work Step by Step

Rewrite the equations so that the variables are on one side while the constant is on the other: $-2x - 5y = -11$ $ 9x + 5y = -3$ We see that in the two equations, the $y$ term is 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: $(-2x-5y)+(9x+5y)=-11+(-3)\\ 7x = -14$ Divide each side by $7$ to solve for $x$: $x = -2$ Now that we have the value for $x$, we can plug it into one of the equations to solve for $y$. Let's plug in the value for $x$ into the first equation: $11 - 5y = 2(-2)$ $11 - 5y = -4$ Now, we subtract $11$ from both sides of the equation to isolate constants to the right side of the equation: $-5y = -15$ Divide both sides by $-5$ to solve for $y$: $y = 3$ The solution to this system of equations is $(-2, 3)$.
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