Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 9



Work Step by Step

Re-write the second equation as: $y=2x-1$ First equation yields: $x(2x-1)=6$ or, $2x^2-x-6=0$ or, $(2x+3)(x-2)=0$ or, $x=${$-\dfrac{3}{2},2$} From first equation when $x=-\dfrac{3}{2}$, we have $y=-4$ From first equation when $x=2$, we have $y=3$ Thus, solution set is $(x,y)=${$(-\dfrac{3}{2},-4),(2,3)$}
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