## Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole

# Chapter 3 - Section 3.1 - Systems of Two Equations in Two Unknowns - Exercises - Page 185: 7

#### Answer

$( \displaystyle \frac{5}{3},-\frac{4}{3})$

#### Work Step by Step

To eliminate y, multiply both equations with numbers so the coefficient of y is $\pm 2:$ $\left\{\begin{array}{llll} 0.5x & +0.1y & =0.7 & /\times 20\\ 0.2x & -0.2y & =0.6 & /\times 10 \end{array}\right.$ $\left\{\begin{array}{llll} 10x & +2y & =14 & \\ 2x & -2y & =6 & \end{array}\right.$ ... add and solve ... $12x=20\qquad/\div 7$ $x=\displaystyle \frac{20}{12}=\frac{5}{3}$ Back substitute into one of the above equations: $2x-2y=6$ $2\displaystyle \cdot\frac{5}{3}-2y=6\qquad /\times 3$ $10-6y=18\qquad /-10$ $-6y=8\qquad/\div(-6)$ $y=-\displaystyle \frac{8}{6}=-\frac{4}{3}$ Solutions are ordered pairs (x,y): $( \displaystyle \frac{5}{3},-\frac{4}{3})$ Check graphically by graphing both lines in the same window and determining the intersection (see image).

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