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: 32

Answer

\[x=2,\ y=-1\]

Work Step by Step

Multiply \[5\]on both sides to the equation \[3x-7y=13\]to get: \[15x-35y=65\]. Multiply \[7\]on both sides to the equation \[6x+5y=7\]to get: \[42x+35y=49\]. Add the above obtained equation from both RHS and LHS as follows: \[\underline{\begin{align} & 15x-35y=65 \\ & 42x-35y=49 \end{align}}\] \[\begin{align} & 57x\ \ \ \ \ \ \ \ \ \ =114 \\ & x=2 \end{align}\] Put \[x=2\]in\[3x-7y=13\], to get: \[\begin{align} & 3\left( 2 \right)-7y=13 \\ & 6-7y=13 \\ & -7y=7 \\ & y=-1 \end{align}\] Put\[x=2\]and \[y=-1\]in any of the given equations to check the solution: \[\begin{align} & 6\left( 2 \right)+5\left( -1 \right)=7 \\ & 12-5=7 \\ & 7=7 \end{align}\] Since RHS\[=\]LHS, it implies the solution is correct. Now, check with other equation 3x – 7y = 13 too. 3(2) -7(-1) = 13 6 +7 = 13 13 = 13 LHS = RHS.So, this is the solution to this system of equations.
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