# Chapter 4 - Section 4.3 - Solving Systems of Linear Equations by Addition - Exercise Set - Page 308: 38

$(\frac{5}{6}, -\frac{1}{2})$

#### Work Step by Step

$3x+\frac{7}{2}y=\frac{3}{4}$ $-\frac{x}{2}+\frac{5}{3}y=-\frac{5}{4}$ Multiply first equation by 4 and second equation by 12. $12x+14y=3$ $-6x+20y=-15$ Multiply the second equation by 2. $12x+14y=3$ $-12x+40y=-30$ Add both equations. $54y=-27$ $y=-\frac{1}{2}$ Solve for $x$. $3x+\frac{7}{2}y=\frac{3}{4}$ $3x+\frac{7}{2}(-\frac{1}{2})=\frac{3}{4}$ $3x-\frac{7}{4}=\frac{3}{4}$ $12x-7=3$ $12x=10$ $x=\frac{5}{6}$ $x=\frac{5}{6}$ $y=-\frac{1}{2}$ $(\frac{5}{6}, -\frac{1}{2})$

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