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: 32

Answer

$x=\dfrac{1}{5}, y = \dfrac{3}{10} \text{ or } \left(\dfrac{1}{5}, \dfrac{3}{10}\right)$

Work Step by Step

The system of equations is given by the following two equations: $$\begin{cases} 3x-2y=0\hspace{20pt} (1)\\[3mm] 5x+10y=4 \hspace{20pt} (2) \end{cases}$$ Multiplying equation $(1)$ by $5$ so that the coefficients of y in the two equations are additive inverses gives: $15x-10y=0\hspace{20pt} (1)$ $\text{Add the two equations to eliminate } y$: \[ \begin{aligned} 15x-10y&=0 \\[2mm] 5x+10y&=4 \ \\[3mm] \hline 20x&=4\\[2mm] \end{aligned} \] Divdei both sides by $20$ to obtain: $x=\dfrac{4}{20}$ $x=\dfrac{1}{5}$ $\text{Substituting in equation (2) for }x=\dfrac{1}{5}$ gives: $5\left(\dfrac{1}{5} \right)+10y = 4$ $1+10y=4$ $10y = 3$ $y = \dfrac{3}{10}$ Therefore, the solution is: $\boxed{x=\dfrac{1}{5} \hspace{20pt} y = \dfrac{3}{10}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.