Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 8 - Systems of Linear Equations and Problem Solving - Connecting: The Concepts - Exercises - Page 516: 6

Answer

$(3,1)$

Work Step by Step

None of the variables have a coefficient 1, meaning that isolating one would involve fractions. The easier way to go here is elimination. Focus on y, which has coefficients with opposite signs. Multiply each equation (don't forget: both sides) as denoted: $\left\{\begin{array}{llll} 2x & -5y & =1 & .../\times 2\\ 3x & +2y & =11 & .../\times 5 \end{array}\right.$ $\left\{\begin{array}{llll} 4x & -10y & =2 & ...\\ 15x & +10y & =55 & ... \end{array}\right.$Adding eliminates y. $19x=57\qquad$ ... and solve for x $x=\displaystyle \frac{57}{19}$ $x=3$ Now, back-substitute into (*) $3x+2y=11$ $ 3(3)+2y=11\qquad$ ... subtract $9$ $2y=2$ $y=1$ Form an ordered pair $(x,y) $ as the solution to the system $(3,1)$
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