Q64. - Algebra 1

Question

Use an augmented matrix to solve each system of equations.

$\begin{array}{l}x-2y=-16\\ 3x+4y=72\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\left(8,12\right)$.

Step 1. Write the augmented matrix.

To write the equations in augmented matrix place the coefficients of the equations and the constant terms into a matrix separated by a dashed line.

 

Here, the augmented matrices are:

 

$\left[\begin{array}{ccc}1& -2& -16\\ 3& 4& 72\end{array}\right]$

Step 2. Use row operations to solve the system.

To make the first element in the second row a 0, multiply the first row by 3 and then subtract resultant row 1 from the row 2.

 

$\left[\begin{array}{ccc}1& -2& -16\\ 3& 4& 72\end{array}\right]\stackrel{R_2\to R_2-3R_1}{\to }\left[\begin{array}{ccc}1& -2& -16\\ 0& 10& 120\end{array}\right]$

 

To make the second element in the second row a 1, divide the second row by 10.

 

 $\left[\begin{array}{ccc}1& -2& -16\\ 0& 10& 120\end{array}\right]\stackrel{\frac{1}{10}{R}_2}{\to }\left[\begin{array}{ccc}1& -2& -16\\ 0& 1& 12\end{array}\right]$

Step 3. Row reduce the matrix.

Further row-reduce the augmented matrix, by making the second element in the first row a zero.

 

$\left[\begin{array}{ccc}1& -2& -16\\ 0& 1& 12\end{array}\right]\stackrel{R_1\to R_1+2R_2}{\to }\left[\begin{array}{ccc}1& 0& 8\\ 0& 1& 12\end{array}\right]$ 

 

Here, the first row will give the solution of x, because the coefficient of y is 0 and the coefficient of x is 1. Similarly, the second row will give the solution of y, because the coefficient of x is 0 and the coefficient of y is 1. Therefore, the solution is $\left(8,12\right)$.