Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 818: 41


The system of equations has a unique solution and the solution is $\left[ \left. \left( x,y \right) \right|\left( \frac{29}{22},-\frac{5}{11} \right) \right]$.

Work Step by Step

Let us consider system of equations: $\begin{align} & 2x-3y=4 \\ & 4x+5y=3 \\ \end{align}$ So, for the equations ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$. If $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$, then the system of equations has a unique solution. Here, ${{a}_{1}}=2,\text{ }{{a}_{2}}=4,\text{ }{{b}_{1}}=-3,\text{ and }{{b}_{2}}=5$. Therefore, $\begin{align} & \frac{2}{4}\ne \frac{-3}{5} \\ & \frac{1}{2}\ne -\frac{3}{5} \end{align}$ Therefore, the system of equations has a unique solution. Multiply the first equation $2x-3y=4$ by -2 to obtain the coefficient of x equal to -4, so that the variable x is eliminated. $-4x+6y=-8$ And add equations $2x=3y+4\text{ and }4x=3-5y$ to obtain: $\begin{align} & 4x+5y-4x+6y=3-8 \\ & 11y=-5 \\ & y=-\frac{5}{11} \end{align}$ Put the value $y=-\frac{5}{11}$ y in the equation $y=-\frac{5}{11}$ to obtain: $\begin{align} & 4x-\frac{25}{11}=3 \\ & 4x=3+\frac{25}{11} \\ & 4x=\frac{58}{11} \\ & x=\frac{29}{22} \end{align}$
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