Finite Math Examples

$2 x + y = 810$ , $x + y = 510$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 1 & 810 \\ 1 & 1 & 510 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{1}{2} & \frac{810}{2} \\ 1 & 1 & 510 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 405 \\ 1 & 1 & 510 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 405 \\ 1 - 1 & 1 - \frac{1}{2} & 510 - 405 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 405 \\ 0 & \frac{1}{2} & 105 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & 405 \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 \cdot 105 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & 405 \\ 0 & 1 & 210 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 405 - \frac{1}{2} \cdot 210 \\ 0 & 1 & 210 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 300 \\ 0 & 1 & 210 \end{array}]$

Step 3

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

$x = 300$

$y = 210$

Step 4

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

$(300, 210)$