Finite Math Examples

$- 16 x \cdot 1 + 12 x \cdot 2 = 4$ , $12 x \cdot 1 - 9 x \cdot 2 = - 3$

Step 1

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

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply $- 16$ by $1$ .

$- 16 x + 12 x \cdot 2 = 4$

$12 x \cdot 1 - 9 x \cdot 2 = - 3$

Step 1.1.2

Multiply $2$ by $12$ .

$- 16 x + 24 x = 4$

$12 x \cdot 1 - 9 x \cdot 2 = - 3$

$- 16 x + 24 x = 4$

$12 x \cdot 1 - 9 x \cdot 2 = - 3$

Step 1.2

Add $- 16 x$ and $24 x$ .

$8 x = 4$

$12 x \cdot 1 - 9 x \cdot 2 = - 3$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $12$ by $1$ .

$8 x = 4$

$12 x - 9 x \cdot 2 = - 3$

Step 1.3.2

Multiply $2$ by $- 9$ .

$8 x = 4$

$12 x - 18 x = - 3$

$8 x = 4$

$12 x - 18 x = - 3$

Step 1.4

Subtract $18 x$ from $12 x$ .

$8 x = 4$

$- 6 x = - 3$

$8 x = 4$

$- 6 x = - 3$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 8 & 4 \\ - 6 & - 3 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{8}{8} & \frac{4}{8} \\ - 6 & - 3 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} \\ - 6 & - 3 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} \\ - 6 + 6 \cdot 1 & - 3 + 6 (\frac{1}{2}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

Step 4

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

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

$0 = 0$

Step 5

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

$(error = error)$