## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 3 - Section 3.4 - Matrix Solutions to Linear Systems - Exercise Set - Page 229: 22

#### Answer

No solution or $\varnothing$.

#### Work Step by Step

The given system of equations is $-3x+4y=12$ $6x-8y=16$ The augmented matrix is $\Rightarrow \left[\begin{array}{cc|c} -3 & 4 & 12\\ 6 & -8 & 16 \end{array}\right]$ Perform $R_1\rightarrow \frac{R_1}{(-3)}$. $\Rightarrow \left[\begin{array}{cc|c} -3/(-3) & 4/(-3) & 12/(-3)\\ 6 & -8 & 16 \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{cc|c} 1 & -4/3 & -4\\ 6 & -8 & 16 \end{array}\right]$ Perform $R_2\rightarrow R_2-6\times R_1$. $\Rightarrow \left[\begin{array}{cc|c} 1 & -4/3 & -4\\ 6-6\times 1 & -8-6\times (-4/3) & 16-6\times (-4) \end{array}\right]$ Simplify. $\Rightarrow \left[\begin{array}{cc|c} 1 & -4/3 & -4\\ 0 & 0 & 40 \end{array}\right]$ Use back substitution to solve the linear system. $\Rightarrow (1)x+(−4/3)y=-4$ and $\Rightarrow (0)x+(0)y=40$ In the second equation no values of $x$ and $y$ satisfy. Hence, the system is inconsistent and has no solution. The solution set is $\varnothing$.

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