## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 10: 19

#### Answer

There is one solution for these equations: (5, -2), or x = 5, y = -2.

#### Work Step by Step

Equation 1: $\frac{x + 3}{4} + \frac{y - 1}{3}$ = 1 Equation 2: 2x - y = 12 Let's simplify this first equation. In order to simplify these fractions, we must multiply both sides of the equation by their lowest common denominator, which is 12 for the numbers 3 and 4. 12($\frac{x + 3}{4} + \frac{y - 1}{3}$) = 12(1) $\frac{12(x + 3)}{4} + \frac{12(y - 1)}{3}$ = 12 (distributive property) 3(x + 3) + 4(y - 1) = 12 (simplify numerator and denominator) 3x + 9 + 4y - 4 = 12 (distributive property) 3x + 4y = 7 (combine like terms and simplify) Our new 2 equations are: Equation 1: 3x + 4y = 7 Equation 2: 2x - y = 12 We can easily eliminate the y variable to solve for x by multiplying equation 2 by 4. The +4y and -4y will cancel out in this way. 4(2x - y) = 4(12) 8x - 4y = 48 (distributive property) Equation 1: 3x + 4y = 7 Equation 2: + 8x - 4y = 48 = 11x + 0y = 55 We can eliminate the 0y because it cancels out. This leaves us with 11x = 55, and dividing both sides by 11 allows us to know that x = 5 (55 / 11 = 5). To solve for y, plug the x value (5) into either of the original equations. 2(5) - y = 12 (plug in) 10 - y = 12 (subtract 10 from both sides) -y = 2 y = -2 (divide each side by -1 to make y positive) The solution is (5, -2), or x = 5, y = -2.

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