Finite Math Examples

\frac{1}{2} x - y = - 3

$\frac{1}{2} x - y = - 3$ ,

9 x - y = 1

$9 x - y = 1$

Step 1

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

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Step 1.1

Combine

\frac{1}{2}

$\frac{1}{2}$ and

x

$x$ .

\frac{x}{2} - y = - 3

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

9 x - y = 1

$9 x - y = 1$

Step 1.2

Reorder terms.

\frac{1}{2} x - y = - 3

$\frac{1}{2} x - y = - 3$

9 x - y = 1

$9 x - y = 1$

\frac{1}{2} x - y = - 3

$\frac{1}{2} x - y = - 3$

9 x - y = 1

$9 x - y = 1$

Step 2

Write the system as a matrix.

[\begin{matrix} \frac{1}{2} & - 1 & - 3 9 & - 1 & 1 \end{matrix}]

$[\begin{array}{c} \frac{1}{2} & - 1 & - 3 \\ 9 & - 1 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of

R_{1}

$R_{1}$ by

2

$2$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 3.1.1

Multiply each element of

R_{1}

$R_{1}$ by

2

$2$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} 2 (\frac{1}{2}) & 2 \cdot - 1 & 2 \cdot - 3 9 & - 1 & 1 \end{matrix}]

$[\begin{array}{c} 2 (\frac{1}{2}) & 2 \cdot - 1 & 2 \cdot - 3 \\ 9 & - 1 & 1 \end{array}]$

Step 3.1.2

Simplify

R_{1}

$R_{1}$ .

$[\begin{array}{c} 1 & - 2 & - 6 \\ 9 & - 1 & 1 \end{array}]$

Step 3.2

Perform the row operation

$R_{2} = R_{2} - 9 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 3.2.1

Perform the row operation

$R_{2} = R_{2} - 9 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - 2 & - 6 \\ 9 - 9 \cdot 1 & - 1 - 9 \cdot - 2 & 1 - 9 \cdot - 6 \end{array}]$

Step 3.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - 6 \\ 0 & 17 & 55 \end{array}]$

Step 3.3

Multiply each element of

$R_{2}$ by

$\frac{1}{17}$ to make the entry at

$2, 2$ a

$1$ .

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Step 3.3.1

Multiply each element of

$R_{2}$ by

$\frac{1}{17}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & - 2 & - 6 \\ \frac{0}{17} & \frac{17}{17} & \frac{55}{17} \end{array}]$

Step 3.3.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - 6 \\ 0 & 1 & \frac{55}{17} \end{array}]$

Step 3.4

Perform the row operation

$R_{1} = R_{1} + 2 R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 3.4.1

Perform the row operation

$R_{1} = R_{1} + 2 R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 6 + 2 (\frac{55}{17}) \\ 0 & 1 & \frac{55}{17} \end{array}]$

Step 3.4.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{8}{17} \\ 0 & 1 & \frac{55}{17} \end{array}]$

Step 4

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

$x = \frac{8}{17}$

$y = \frac{55}{17}$

Step 5

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

$(\frac{8}{17}, \frac{55}{17})$

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