Chapter 6.7, Problem 45PPS

To determine

To solve: The given system of equation

Answer to Problem 45PPS

Here the solution to the system of equation is $\left(\frac{-2}{3}+\frac{1}{3}t,t\right)$, where t is any arbitrary real

Explanation of Solution

Given: $\begin{array}{l}6x-2y=-4\\ -3x+y=2\end{array}$

Calculation:

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

Place the coefficient of the equation and the constant term into a matrix

  $\begin{array}{l}6x-2y=-4\\ -3x+y=2\end{array}$

  $\left[\begin{array}{l}\text{6 }-\text{2 | }-\text{4}\\ -\text{3 1 }\text{| 2}\end{array}\right]$

Therefore, augmented matrix is

  $\left[\begin{array}{l}\text{6 }-\text{2 | }-\text{4}\\ -\text{3 1 }\text{| 2}\end{array}\right]$

To make the first element in the first row a $1$ , multiply first row by $\frac{1}{6}$

  $\left[\begin{array}{l}\text{6 }-\text{2 | }-\text{4}\\ -\text{3 1 }\text{| 2}\end{array}\right]$

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

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

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

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

This gives $x+\frac{-1}{3}y=\frac{-2}{3}\Rightarrow x=\frac{-2}{3}+\frac{1}{3}y$

Hence the solution to the system of equation is $\left(\frac{-2}{3}+\frac{1}{3}t,t\right)$ , where t is any arbitrary real number.

Conclusion:

Therefore, the solution to the system of equation is $\left(\frac{-2}{3}+\frac{1}{3}t,t\right)$, where t is any arbitrary real