Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 7 - Algebra: Graphs, Functions, and Linear Systems - 7.3 Systems of Linear Equations in Two Variables - Exercise Set 7.3 - Page 444: 31

Answer

\[x=3,\ y=1\]

Work Step by Step

Multiply \[5\]on both sides to the equation \[4x+3y=15\]to get: \[20x+15y=75\]. Multiply \[3\]on both sides to the equation \[2x-5y=1\]to get: \[6x-15y=3\]. Add the above obtained equation from both RHS and LHS as follows: \[\underline{\begin{align} & 20x+15y=75 \\ & 6x-15y=3 \end{align}}\] \[\begin{align} & 26x\ \ \ \ \ \ \ \ =78 \\ & x=3 \end{align}\] Put \[x=3\]in\[4x+3y=15\], to get: \[\begin{align} & 4\left( 3 \right)+3y=15 \\ & 12+3y=15 \\ & 3y=3 \\ & y=1 \end{align}\] Put\[x=3\]and \[y=1\]in any of the given equations to check the solution: \[\begin{align} & 2\left( 3 \right)-5\left( 1 \right)=1 \\ & 6-5=1 \\ & 1=1 \end{align}\] Since RHS\[=\]LHS, it implies the solution is correct. Now check with 4x + 3y = 15 too: 4(3) + 3(1) = 15 12 + 3 = 15 15 = 15 LHS = RHS Above solution satisfies both the equations.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.