Algebra: A Combined Approach (4th Edition)

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

Chapter 4 - Test - Page 331: 15

Answer

$$x=5$$ $$y=-2$$

Work Step by Step

Equation 1: $\frac{x-3}{2}=\frac{2-y}{4}$ Equation 2: $\frac{7-2x}{3}=\frac{y}{2}$ Simplify the equations by cross-multiplying the numerator of the left-hand side with the denominator of the right-hand side, and the numerator of the right-hand side with the denominator of the left-hand side. Equation 1: $$\frac{x-3}{2}=\frac{2-y}{4}$$ $$(x-3)(4)=(2-y)(2)$$ $$4x-12=4-2y$$ We can call this equation 1'. Equation 2: $$\frac{7-2x}{3}=\frac{y}{2}$$ $$(7-2x)(2)=3(y)$$ $$14-4x=3y$$ We can call this equation 2'. Using equation 2, divide both sides by $3$: $$14-4x=3y$$ $$\frac{14-4x}{3}=\frac{3y}{3}$$ $$\frac{14-4x}{3}=y$$ Substitute this equation to equation 1': $$4x-12=4-2(\frac{14-4x}{3})$$ $$4x-12=4-(\frac{28-8x}{3})$$ Multiply the whole equation by $3$: $$[4x-12=4-(\frac{28-8x}{3})]\cdot3$$ $$12x-36=12-(28-8x)$$ $$12x-36=12-28+8x$$ $$12x-36=-16+8x$$ Add $36$ to both sides: $$12x-36=-16+8x$$ $$12x-36+36=-16+8x+36$$ $$12x=8x+20$$ Subtract $8x$ from both sides: $$12x=8x+20$$ $$12x-8x=8x-8x+20$$ $$4x=20$$ Divide both sides by $4$: $$\frac{4x}{4}=\frac{20}{4}$$ $$x=5$$ Substitute this value of $x$ to equation 1': $$4x-12=4-2y$$ $$4(5)-12=4-2y$$ $$20-12=4-2y$$ $$8=4-2y$$ Subtract $4$ from both sides: $$8-4=4-4-2y$$ $$4=-2y$$ Divide both sides by $-2$: $$\frac{4}{-2}=\frac{-2y}{-2}$$ $$-2=y$$ $$y=-2$$ Check using equation 2': $$14-4x=3y$$ $$14-4(5)=3(-2)$$ $$14-20=-6$$ $$-6=-6$$
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