Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 818: 25


The required value is $x=3,y=1$

Work Step by Step

We have to multiply the second equation with 2 and then subtract this from the first equation: $\begin{align} & 4x+3y-2\left( 2x-5y \right)=15-2\left( 1 \right) \\ & 4x+3y-4x+10y=15-2 \\ & 13y=13 \\ & y=1 \end{align}$ Substitute the value of $y=1$ in the second equation: $\begin{align} & 2x-5\left( 1 \right)=1 \\ & 2x=1+5 \\ & 2x=6 \\ & x=3 \end{align}$ Thus, the value of $x=3$ and $y=1$.
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