Chapter 6.7, Problem 18PPS

To determine

To solve: the given system.

Answer to Problem 18PPS

The solution to the system of equation is $\left(7,3\right)$

Explanation of Solution

Given:

  $\begin{array}{l}x-3y=-2\\ 4x+y=31\end{array}$

Calculation:

Objective: To use the augmented matrix to solve the given system.

Place the coefficients of the equations and the constant term into a matrix

  $\begin{array}{l}x-3y=-2\\ 4x+y=31\end{array}$

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{4 1 | 31}\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{4 1 | 31}\end{array}\right]$

To make the first element in the second row a $0$ , multiply first row by $-4$ and add the result to the second row

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{4 1 | 31}\end{array}\right]$

  $\left(-4\right)R_1+R_2$

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{0 13 | 39}\end{array}\right]$

To make the second element in the second row a $1$ , multiply second row by $\frac{1}{13}$

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{0 13 | 39}\end{array}\right]$

  $\frac{1}{13}{R}_2$

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{0 1 | 3}\end{array}\right]$

To make the second element in the first row a $0$ , multiply second row by $3$ and add the result to the first row

  $\left[\begin{array}{l}1\text{}\text{ -3 | -2}\\ \text{0 1 | 3}\end{array}\right]$

  $R_1+3R_2$

  $\left[\begin{array}{l}1\text{}\text{ 0 | 7}\\ \text{0 1 | 3}\end{array}\right]$

Hence the solution to the system of equation is $\left(7,3\right)$ .

Conclusion:

The solution to the system of equation is $\left(7,3\right)$