Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 1, Equations and Inequalities - 1.3 Solve Linear Equations - 1.3 Exercises - Skill Practice - Page 23: 67


there is no solution when $a=c$ but $d\ne b$. all real numbers If $a=c$ and $d=b$.

Work Step by Step

$$ax+b = cx+d$$ $$ax-cx = d-b$$ $$(a-c)x = d-b$$ $$x = \frac{d-b}{a-c}$$ Thus there is no solution when $a=c$ but $d\ne b$. If $a=c$ and $d=b$, $ax+b = ax+b$, which is always true.
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