Finite Math Examples

$4 x - 2 y = - 10$ , $2 x + 4 y = 0$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 4 & - 2 & - 10 \\ 2 & 4 & 0 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 2}{4} & \frac{- 10}{4} \\ 2 & 4 & 0 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{2} \\ 2 & 4 & 0 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{2} \\ 2 - 2 \cdot 1 & 4 - 2 (- \frac{1}{2}) & 0 - 2 (- \frac{5}{2}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{2} \\ 0 & 5 & 5 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{2} \\ \frac{0}{5} & \frac{5}{5} & \frac{5}{5} \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{2} \\ 0 & 1 & 1 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & - \frac{5}{2} + \frac{1}{2} \cdot 1 \\ 0 & 1 & 1 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

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

Step 3

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

$x = - 2$

$y = 1$

Step 4

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

$(- 2, 1)$