Finite Math Examples

$(2 x + 2 y) = 4$ , $0.5 x + y = 1$

Step 1

Remove parentheses.

$2 x + 2 y = 4$

$0.5 x + y = 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 2 & 2 & 4 \\ 0.5 & 1 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{2}{2} & \frac{4}{2} \\ 0.5 & 1 & 1 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 2 \\ 0.5 & 1 & 1 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 0.5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 0.5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 2 \\ 0.5 - 0.5 \cdot 1 & 1 - 0.5 \cdot 1 & 1 - 0.5 \cdot 2 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 2 \\ 0 & 0.5 & 0 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{1}{0.5}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{0.5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 2 \\ \frac{0}{0.5} & \frac{0.5}{0.5} & \frac{0}{0.5} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 2$

$y = 0$

Step 5

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

$(2, 0)$