Finite Math Examples

$4 x - 72 = 72$ , $5 x - 8 y = - 8$

Step 1

Move all terms not containing a variable to the right side of the equation.

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Step 1.1

Add $72$ to both sides of the equation.

$4 x = 72 + 72$

$5 x - 8 y = - 8$

Step 1.2

Add $72$ and $72$ .

$4 x = 144$

$5 x - 8 y = - 8$

$4 x = 144$

$5 x - 8 y = - 8$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 4 & 0 & 144 \\ 5 & - 8 & - 8 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{0}{4} & \frac{144}{4} \\ 5 & - 8 & - 8 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 36 \\ 5 & - 8 & - 8 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 36 \\ 5 - 5 \cdot 1 & - 8 - 5 \cdot 0 & - 8 - 5 \cdot 36 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 36 \\ 0 & - 8 & - 188 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & 36 \\ - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot - 8 & - \frac{1}{8} \cdot - 188 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 36 \\ 0 & 1 & \frac{47}{2} \end{array}]$

Step 4

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

$x = 36$

$y = \frac{47}{2}$

Step 5

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

$(36, \frac{47}{2})$