Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 4 - Review - Page 328: 18

Answer

True for all values of $x$ and $y$.

Work Step by Step

equation 1 $$9x = 6y + 3$$ equation 2 $$6x - 4y = 2$$ Using equation 1: Divide both sides by $9$: $$9x = 6y + 3$$ $$\frac{9x}{9} = \frac{6y + 3}{9}$$ $$x = \frac{6y + 3}{9}$$ We will refer to this equation as equation 1' Substituting this to equation 2: $$6x - 4y = 2$$ $$6(\frac{6y + 3}{9}) - 4y = 2$$ $$\frac{36y + 18}{9} - 4y = 2$$ Multiply both sides by $9$: $$ 9\cdot[\frac{36y + 18}{9} - 4y] = 2\cdot9$$ $$36y + 18-36y=18$$ $$36y+18=36y + 18$$ Since the expression is true, then the equation is true for all values of $y$. Since the system is true for all values of $y$, then it also has to be true for all values of $x$ in order to satisfy the equations.
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