Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 8 - Systems of Linear Equations and Problem Solving - 8.2 Solving by Substitution or Elimination - 8.2 Exercise Set - Page 516: 17


The solution set is $\{(t, 2t-3)\ |\ t\in \mathbb{R}\}$

Work Step by Step

Variable $y$ is isolated in eq.$2$ $y=2x-3\qquad $(*) Replace $y$ with $2x-3$ in the other equation: $ 4x-2(2x-3)=6\qquad$ ... and solve for $x$. Simplify $4x-4x+6=6$ $ 0=0\qquad$ ... always true, there are infinitely many solutions This is the case where both equations have graphs that coincide. All points on that common line have coordinates that satisfy both equations. Letting $x=t\in \mathbb{R}$ (be any real number), from equation $(*)$ it follows: $y=2t-3$ The solution set is $\{(t, 2t-3)\ |\ t\in \mathbb{R}\}$
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