Finite Math Examples

- 4 x + 5 y = 13

$- 4 x + 5 y = 13$ ,

- 12 x + 15 y = 39

$- 12 x + 15 y = 39$

Step 1

Write the system as a matrix.

[\begin{matrix} - 4 & 5 & 13 - 12 & 15 & 39 \end{matrix}]

$[\begin{array}{c} - 4 & 5 & 13 \\ - 12 & 15 & 39 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of

R_{1}

$R_{1}$ by

- \frac{1}{4}

$- \frac{1}{4}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 2.1.1

Multiply each element of

R_{1}

$R_{1}$ by

- \frac{1}{4}

$- \frac{1}{4}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 5 & - \frac{1}{4} \cdot 13 - 12 & 15 & 39 \end{matrix}]

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 5 & - \frac{1}{4} \cdot 13 \\ - 12 & 15 & 39 \end{array}]$

Step 2.1.2

Simplify

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & - \frac{5}{4} & - \frac{13}{4} - 12 & 15 & 39 \end{matrix}]

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{13}{4} \\ - 12 & 15 & 39 \end{array}]$

Step 2.2

Perform the row operation

$R_{2} = R_{2} + 12 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 2.2.1

Perform the row operation

$R_{2} = R_{2} + 12 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{13}{4} \\ - 12 + 12 \cdot 1 & 15 + 12 (- \frac{5}{4}) & 39 + 12 (- \frac{13}{4}) \end{array}]$

Step 2.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{4} & - \frac{13}{4} \\ 0 & 0 & 0 \end{array}]$

Step 3

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

$x - \frac{5}{4} y = - \frac{13}{4}$

$0 = 0$

Step 4

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

$(- \frac{13}{4} + \frac{5 y}{4}, y)$