Finite Math Examples

$\frac{1}{2} x - \frac{1}{4} y = 1$ , $\frac{1}{3} x + y = 3$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Simplify each term.

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Step 1.1.1

Combine $\frac{1}{2}$ and $x$ .

$\frac{x}{2} - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

Step 1.1.2

Combine $y$ and $\frac{1}{4}$ .

$\frac{x}{2} - \frac{y}{4} = 1$

$\frac{1}{3} x + y = 3$

$\frac{x}{2} - \frac{y}{4} = 1$

$\frac{1}{3} x + y = 3$

Step 1.2

Reorder terms.

$\frac{1}{2} x - (\frac{1}{4} y) = 1$

$\frac{1}{3} x + y = 3$

Step 1.3

Remove parentheses.

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

Step 1.4

Combine $\frac{1}{3}$ and $x$ .

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{x}{3} + y = 3$

Step 1.5

Reorder terms.

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

$\frac{1}{2} x - \frac{1}{4} y = 1$

$\frac{1}{3} x + y = 3$

Step 2

Write the system as a matrix.

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{4} & 1 \\ \frac{1}{3} & 1 & 3 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 2 (\frac{1}{2}) & 2 (- \frac{1}{4}) & 2 \cdot 1 \\ \frac{1}{3} & 1 & 3 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 2 \\ \frac{1}{3} & 1 & 3 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - \frac{1}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{3} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 2 \\ \frac{1}{3} - \frac{1}{3} \cdot 1 & 1 - \frac{1}{3} (- \frac{1}{2}) & 3 - \frac{1}{3} \cdot 2 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 2 \\ 0 & \frac{7}{6} & \frac{7}{3} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{6}{7}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{6}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 2 \\ \frac{6}{7} \cdot 0 & \frac{6}{7} \cdot \frac{7}{6} & \frac{6}{7} \cdot \frac{7}{3} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

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

Step 3.4

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 3.4.1

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

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

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 3 \\ 0 & 1 & 2 \end{array}]$

Step 4

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

$x = 3$

$y = 2$

Step 5

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

$(3, 2)$