Finite Math Examples

$\frac{1}{3} x + \frac{1}{2} y = 7$ , $\frac{2}{3} x - \frac{5}{2} y = - 14$

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

$\frac{x}{3} + \frac{1}{2} y = 7$

$\frac{2}{3} x - \frac{5}{2} y = - 14$

Step 1.1.2

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

$\frac{x}{3} + \frac{y}{2} = 7$

$\frac{2}{3} x - \frac{5}{2} y = - 14$

$\frac{x}{3} + \frac{y}{2} = 7$

$\frac{2}{3} x - \frac{5}{2} y = - 14$

Step 1.2

Reorder terms.

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2}{3} x - \frac{5}{2} y = - 14$

Step 1.3

Simplify each term.

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Step 1.3.1

Combine $\frac{2}{3}$ and $x$ .

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2 x}{3} - \frac{5}{2} y = - 14$

Step 1.3.2

Combine $y$ and $\frac{5}{2}$ .

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2 x}{3} - \frac{y \cdot 5}{2} = - 14$

Step 1.3.3

Move $5$ to the left of $y$ .

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2 x}{3} - \frac{5 y}{2} = - 14$

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2 x}{3} - \frac{5 y}{2} = - 14$

Step 1.4

Reorder terms.

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2}{3} x - (\frac{5}{2} y) = - 14$

Step 1.5

Remove parentheses.

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2}{3} x - \frac{5}{2} y = - 14$

$\frac{1}{3} x + \frac{1}{2} y = 7$

$\frac{2}{3} x - \frac{5}{2} y = - 14$

Step 2

Write the system as a matrix.

$[\begin{array}{c} \frac{1}{3} & \frac{1}{2} & 7 \\ \frac{2}{3} & - \frac{5}{2} & - 14 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $3$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $3$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 3 (\frac{1}{3}) & 3 (\frac{1}{2}) & 3 \cdot 7 \\ \frac{2}{3} & - \frac{5}{2} & - 14 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 21 \\ \frac{2}{3} & - \frac{5}{2} & - 14 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 21 \\ \frac{2}{3} - \frac{2}{3} \cdot 1 & - \frac{5}{2} - \frac{2}{3} \cdot \frac{3}{2} & - 14 - \frac{2}{3} \cdot 21 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 21 \\ 0 & - \frac{7}{2} & - 28 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 21 \\ - \frac{2}{7} \cdot 0 & - \frac{2}{7} (- \frac{7}{2}) & - \frac{2}{7} \cdot - 28 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 21 \\ 0 & 1 & 8 \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & 21 - \frac{3}{2} \cdot 8 \\ 0 & 1 & 8 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 9 \\ 0 & 1 & 8 \end{array}]$

Step 4

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

$x = 9$

$y = 8$

Step 5

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

$(9, 8)$