Finite Math Examples

$- x - 7 y = 13$ , $4 x + 28 y = - 52$

Step 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 7 & 13 \\ 4 & 28 & - 52 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - - 1 & - - 7 & - 1 \cdot 13 \\ 4 & 28 & - 52 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 7 & - 13 \\ 4 & 28 & - 52 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 7 & - 13 \\ 4 - 4 \cdot 1 & 28 - 4 \cdot 7 & - 52 - 4 \cdot - 13 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 7 & - 13 \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x + 7 y = - 13$

$0 = 0$

Step 4

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

$(- 13 - 7 y, y)$