Algebra 2 Common Core

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

Chapter 3 - Linear Systems - 3-3 Systems of Inequalities - Practice and Problem-Solving Exercises - Page 155: 66

Answer

The solution to this system of equations is $(-2, -1)$.

Work Step by Step

It looks like using the elimination method is an easier way to solve this equation. First, we need to convert the equations so that one variable has opposite numerical coefficients. If we add these two equations together, we can eliminate one variable and just deal with one variable instead of two: It is easier to get rid of the $y$ terms in both equations because the $y$ terms are already exactly the same, including the sign, so we want to multiply the first equation by $-1$ to change the sign of the $y$ term in the first equation to be able to eliminate the $x$ term: $-2x - 4y = 8$ $-5x + 4y = 6$ Add the two equations together to obtain: $(-2x-4y)+(-5x+4y)=8+6\\ -7x = 14$ 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 us plug in the value for $x$ into the first equation: $2x+4y=-8\\ 2(-2) + 4y = -8$ $-4 + 4y = -8$ Add $4$ to both sides of the equation to isolate constants to the right side of the equation: $4y = -4$ Divide both sides by $4$ to solve for $y$: $y = -1$ The solution to this system of equations is $(-2, -1)$.
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