Finite Math Examples

$7 x + 4 y = 8$ , $y = - \frac{7}{4} x + 2$

Step 1

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

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Step 1.1

Add $\frac{7}{4} x$ to both sides of the equation.

$7 x + 4 y = 8$

$y + \frac{7}{4} x = 2$

Step 1.2

Combine $\frac{7}{4}$ and $x$ .

$7 x + 4 y = 8$

$y + \frac{7 x}{4} = 2$

Step 1.3

Reorder terms.

$7 x + 4 y = 8$

$y + \frac{7}{4} x = 2$

Step 1.4

Reorder $y$ and $\frac{7}{4} x$ .

$7 x + 4 y = 8$

$\frac{7}{4} x + y = 2$

$7 x + 4 y = 8$

$\frac{7}{4} x + y = 2$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 7 & 4 & 8 \\ \frac{7}{4} & 1 & 2 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{7}{7} & \frac{4}{7} & \frac{8}{7} \\ \frac{7}{4} & 1 & 2 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{4}{7} & \frac{8}{7} \\ \frac{7}{4} & 1 & 2 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - \frac{7}{4} R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{4} R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{7} & \frac{8}{7} \\ \frac{7}{4} - \frac{7}{4} \cdot 1 & 1 - \frac{7}{4} \cdot \frac{4}{7} & 2 - \frac{7}{4} \cdot \frac{8}{7} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{7} & \frac{8}{7} \\ 0 & 0 & 0 \end{array}]$

Step 4

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

$x + \frac{4}{7} y = \frac{8}{7}$

$0 = 0$

Step 5

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

$(\frac{8}{7} - \frac{4 y}{7}, y)$