Math Problem Solver

The system of linear equations has exactly one solution.

\[x = 8\]

Begin by writing the augmented matrix of the system of equations. $$\begin{bmatrix}-7 & -5 &\bigm |& -116\\0 & 1 &\bigm |& 12\end{bmatrix}$$

Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:

  1. Swap the positions of any two rows.

  2. Multiply any row by a nonzero scalar.

  3. Multiply a row by a nonzero scalar and add it to any other row.

First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.

$$\begin{bmatrix}-7 & -5 &\bigm |& -116\\0 & 1 &\bigm |& 12\end{bmatrix}$$

Multiply row \(1\) by scalar \(- \frac{1}{7}\) to make the leading term \(1\).

$$\begin{bmatrix}1 & \frac{5}{7} &\bigm |& \frac{116}{7}\\0 & 1 &\bigm |& 12\end{bmatrix}$$

The leading term of row \(2\) is already \(1\) so this row does not need to be multiplied by a scalar.

$$\begin{bmatrix}1 & \frac{5}{7} &\bigm |& \frac{116}{7}\\0 & 1 &\bigm |& 12\end{bmatrix}$$

Multiply row \(2\) by scalar \(- \frac{5}{7}\) and add it to row \(1\).

$$\begin{bmatrix}1 & 0 &\bigm |& 8\\0 & 1 &\bigm |& 12\end{bmatrix}$$

Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. $$\begin{aligned}1 \cdot x+ 0 \cdot y = 8 \\ x = 8\end{aligned}$$$$\begin{aligned}0 \cdot x+ 1 \cdot y = 12 \\ y = 12\end{aligned}$$