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 - 8.2 Solving by Substitution or Elimination - 8.2 Exercise Set - Page 516: 15

Answer

$(2,-2)$

Work Step by Step

Isolate $t$ in eq.$2$ $ 5s+t=8\qquad$ ... subtract $5s$ $t=8-5s\qquad $(*) Replace $t$ with $8-5s$ in the other equation: $ 3s-4(8-5s)=14\qquad$ ... and solve for $x$. Simplify $3s-32+20s=14$ $ 23s-32=14\qquad$ ... add 32 $ 23s=46\qquad$ ... divide with $23$ $s=2$ Now, back-substitute into (*) $t=8-5(2)=8-10=-2$ Form an ordered pair $(s,t) $ as the solution to the system $(2,-2)$
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