Finite Math Examples

$x - \frac{1}{2} y = 1$ , $\frac{7}{2} x - \frac{1}{7} y = 1$

Step 1

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

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Step 1.1

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

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Step 1.2

Reorder terms.

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Step 1.3

Remove parentheses.

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Step 1.4

Simplify each term.

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Step 1.4.1

Combine $\frac{7}{2}$ and $x$ .

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

$\frac{7 x}{2} - \frac{1}{7} y = 1$

Step 1.4.2

Combine $y$ and $\frac{1}{7}$ .

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

$\frac{7 x}{2} - \frac{y}{7} = 1$

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

$\frac{7 x}{2} - \frac{y}{7} = 1$

Step 1.5

Reorder terms.

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

$\frac{7}{2} x - (\frac{1}{7} y) = 1$

Step 1.6

Remove parentheses.

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

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

$\frac{7}{2} x - \frac{1}{7} y = 1$

Step 2

Write the system as a matrix.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ 0 & \frac{45}{28} & - \frac{5}{2} \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $\frac{28}{45}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $\frac{28}{45}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ \frac{28}{45} \cdot 0 & \frac{28}{45} \cdot \frac{45}{28} & \frac{28}{45} (- \frac{5}{2}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 1 \\ 0 & 1 & - \frac{14}{9} \end{array}]$

Step 3.3

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 3.3.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 1 + \frac{1}{2} (- \frac{14}{9}) \\ 0 & 1 & - \frac{14}{9} \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{2}{9} \\ 0 & 1 & - \frac{14}{9} \end{array}]$

Step 4

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

$x = \frac{2}{9}$

$y = - \frac{14}{9}$

Step 5

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

$(\frac{2}{9}, - \frac{14}{9})$