Chapter 8, Problem 106RE

To determine

Find the value of the given system of linear equation using graphing utility.

Answer to Problem 106RE

The values of the given system of linear equations are $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-\frac{329}{5}\\ \frac{168}{5}\end{array}\right]$ .

Explanation of Solution

Given information:

The given system of linear equations are

  $\begin{array}{l}5x+10y=7\\ 2x+y=-98\end{array}$ .

Calculation:

If $A$ is an invertible matrix, then the system of linear equations represented by $AX=B$ has a unique solution given by $X=A^{-1}B$ .

Write the equation in matrix form $AX=B$ .

Use solving of linear equations calculator.

Select the equations edit menu.

  $\left[\begin{array}{ccc}& & \\ & & \\ & & \end{array}\right]$

Enter the equations in select menu.

  $\left[\begin{array}{l}5x+10y=7\\ 2x+y=-98\end{array}\right]$

Press the ‘solve using the inverse matrix’ key.

The resulting display on the calculator is

  $A=\left[\begin{array}{cc}5& 10\\ 2& 1\end{array}\right]$

  $B=\left[\begin{array}{c}7\\ -98\end{array}\right]$

  $A^{-1}=\left[\begin{array}{cc}-\frac{1}{15}& \frac{2}{3}\\ \frac{12}{15}& -\frac{1}{3}\end{array}\right]$

  $X=A^{-1}\times B=\left[\begin{array}{cc}-\frac{1}{15}& \frac{2}{3}\\ \frac{12}{15}& -\frac{1}{3}\end{array}\right]\left[\begin{array}{c}7\\ -98\end{array}\right]=\left[\begin{array}{c}-\frac{329}{5}\\ \frac{168}{5}\end{array}\right]$

Hence thevalues of the given system of linear equations are $\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-\frac{329}{5}\\ \frac{168}{5}\end{array}\right]$ .