Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 3 - Section 3.6 - Linear Inequalities in Two Variables - Exercise Set - Page 271: 75

Answer

The point $(4, -1)$ is a solution of both equations.

Work Step by Step

To determine whether or not the point $(4, -1)$ is a solution of both equations, we need to substitute $4$ for $x$ and $-1$ for $y$ into both equations to see if both equations are true. Let us plug the values into the first equation to see if it is true: $$4 + 2(-1) = 2$$ Multiply first, according to the order of operations: $$4 + (-2) = 2$$ To add a negative number means to subtract the number, so we rewrite the equation this way: $$4 - 2 = 2$$ Subtract: $$2 = 2$$ The point $(4, -1)$ is a solution of the first equation. Now, let us plug in this point into the second equation to see if the equation is true: $$4 - 2(-1) = 6$$ Multiply first, according to order of operations: $$4 - (-2) = 6$$ To subtract a negative number means to add the number. Let us rewrite the equation to reflect this: $$4 + 2 = 6$$ Do the addition: $$6 = 6$$ The point $(4, -1)$ is also a solution of the second equation, so it is a solution of both equations.
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