Finite Math Examples

$x + 16 y = 9$ , $\frac{1}{4} x + 4 y = 8$

Step 1

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

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Step 1.1

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

$x + 16 y = 9$

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

Step 1.2

Reorder terms.

$x + 16 y = 9$

$\frac{1}{4} x + 4 y = 8$

$x + 16 y = 9$

$\frac{1}{4} x + 4 y = 8$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 16 & 9 \\ \frac{1}{4} & 4 & 8 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 16 & 9 \\ \frac{1}{4} - \frac{1}{4} \cdot 1 & 4 - \frac{1}{4} \cdot 16 & 8 - \frac{1}{4} \cdot 9 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 16 & 9 \\ 0 & 0 & \frac{23}{4} \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $\frac{4}{23}$ to make the entry at $2, 3$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $\frac{4}{23}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & 16 & 9 \\ \frac{4}{23} \cdot 0 & \frac{4}{23} \cdot 0 & \frac{4}{23} \cdot \frac{23}{4} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 16 & 9 \\ 0 & 0 & 1 \end{array}]$

Step 3.3

Perform the row operation $R_{1} = R_{1} - 9 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{1} = R_{1} - 9 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 9 \cdot 0 & 16 - 9 \cdot 0 & 9 - 9 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 16 & 0 \\ 0 & 0 & 1 \end{array}]$

Step 4

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

$x + 16 y = 0$

$0 = 1$

Step 5

The system is inconsistent so there is no solution.

$No solution$