Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.1 Systems of Linear Equations: Substitution and Elimination - 10.1 Assess Your Understanding - Page 734: 33


$x=\dfrac{3}{2}, y = 1 \text{ or } \left(\dfrac{3}{2}, 1\right)$

Work Step by Step

The system of equations is given by the following two equations: $$\begin{cases} 2x+3y&=6\hspace{20pt} &(1)\\[3mm] x-y&=\dfrac{1}{2} \hspace{20pt} &(2) \end{cases}$$ Multiplying equation $(2)$ by $3$ so that the coefficients of y in the two equations are additive inverses gives: $3x-3y=\dfrac{3}{2}\hspace{20pt} (1)$ $\text{Add the two equations to eliminate } y:$ \[ \begin{aligned} 2x+3y&=6 \\[2mm] 3x-3y&=\dfrac{3}{2} \ \\[3mm] \hline 5x&=\dfrac{15}{2}\\[2mm] \end{aligned} \] Divide both sides by $5$ to obtain: $x=\dfrac{\dfrac{15}{2}}{5}$ $x=\dfrac{3}{2}$ $\text{Substituting in equation (2) for }x=\dfrac{3}{2}$ gives: $\dfrac{3}{2}-y=\dfrac{1}{2}$ $y=\dfrac{3}{2}-\dfrac{1}{2}$ $y=1$ Therefore, the solution is: $\boxed{x=\dfrac{3}{2} \hspace{20pt} y = 1}$
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