Finite Math Examples

$x + 7 y = 25$ , $- 7 x + 5 y = - 13$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 7 & 25 \\ - 7 & 5 & - 13 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Perform the row operation $R_{2} = R_{2} + 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.1.1

Perform the row operation $R_{2} = R_{2} + 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 7 & 25 \\ - 7 + 7 \cdot 1 & 5 + 7 \cdot 7 & - 13 + 7 \cdot 25 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 7 & 25 \\ 0 & 54 & 162 \end{array}]$

Step 2.2

Multiply each element of $R_{2}$ by $\frac{1}{54}$ to make the entry at $2, 2$ a $1$ .

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Step 2.2.1

Multiply each element of $R_{2}$ by $\frac{1}{54}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 7 & 25 \\ \frac{0}{54} & \frac{54}{54} & \frac{162}{54} \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 7 & 25 \\ 0 & 1 & 3 \end{array}]$

Step 2.3

Perform the row operation $R_{1} = R_{1} - 7 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{1} = R_{1} - 7 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 7 \cdot 0 & 7 - 7 \cdot 1 & 25 - 7 \cdot 3 \\ 0 & 1 & 3 \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 4 \\ 0 & 1 & 3 \end{array}]$

Step 3

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

$x = 4$

$y = 3$

Step 4

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

$(4, 3)$