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: 31

Answer

Infinite solutions The solutions are the points on the line $y = \frac{2}{3}x + \frac{13}{3}$.

Work Step by Step

Given: $4x - 6y = -26$ $-2x + 3y = 13$ Multiply both sides of the second equation, $-2x + 3y = 13$, by $2$. Gives: $-4x + 6y = 26$ Add the equations, $4x - 6y = -26$ and $-4x + 6y = 26$: $4x - 6y + (-4x + 6y) = -26 + (26) \\4x - 4x - 6y + 6y = -26 + 26$ Combine like terms: $0 = 0$ This statement is true. This implies that Therefore, there are infinite solutions to this problem. Solve for $y$ to find the formula for solutions. Using the first equation: $4x - 6y = -26$ Divide both sides by $2$: $2x - 3y = -13$ Add $3y$ to both sides: $2x = -13 + 3y$ Add $13$ to both sides: $2x + 13 = 3y$ Divide both sides by $3$: $\frac{2}{3}x + \frac{13}{3} = y$ Therefore, the solutions to the given system of linear equations are all the points on the line $y = \frac{2}{3}x + \frac{13}{3}$.
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