Solve Using a Matrix by Elimination x+y=1 2x+5y=4

$x + y = 1$ $2 x + 5 y = 4$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 \\ 2 & 5 & 4 \end{array}]$

Step 2

Find the reduced row echelon form.

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot 1 & 4 - 2 \cdot 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 3 & 2 \end{array}]$

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 \\ \frac{0}{3} & \frac{3}{3} & \frac{2}{3} \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & \frac{2}{3} \end{array}]$

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 1 - \frac{2}{3} \\ 0 & 1 & \frac{2}{3} \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{3} \\ 0 & 1 & \frac{2}{3} \end{array}]$

Step 3

Use the result matrix to declare the final solution to the system of equations.

$x = \frac{1}{3} y = \frac{2}{3}$

Step 4

The solution is the set of ordered pairs that make the system true.

$(\frac{1}{3}, \frac{2}{3})$