Chapter 6.7, Problem 19PPS

To determine

To solve: the given system.

Answer to Problem 19PPS

The solution to the system of equation is $\left(-5,4\right)$

Explanation of Solution

Given:

  $\begin{array}{l}x+2y=3\\ -3x+3y=27\end{array}$

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+2y=3\\ -3x+3y=27\end{array}$$\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{-3 3 | 27}\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{-3 3 | 27}\end{array}\right]$

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

  $\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{-3 3 | 27}\end{array}\right]$

  $3R_1+R_2$$\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{0 9 | 36}\end{array}\right]$

To make the second element in the second row a $1$ , multiply second row by $\frac{1}{9}$

  $\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{0 9 | 36}\end{array}\right]$

  $\frac{1}{9}{R}_2$

  $\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{0 1 | 4}\end{array}\right]$

To make the second element in the first row a $0$ , multiply second row by $-2$ and add the result to the first row

  $\left[\begin{array}{l}1\text{}\text{ 2 | 3}\\ \text{0 1 | 4}\end{array}\right]$

  $R_1+\left(-2\right)R_2$

  $\left[\begin{array}{l}1\text{}\text{ 0 | -5}\\ \text{0 1 | 4}\end{array}\right]$

Hence the solution to the system of equation is $\left(-5,4\right)$ .

Conclusion:

The solution to the system of equation is $\left(-5,4\right)$