Solve Using an Augmented Matrix 6x1+4x2=8 -12x1-8x2=-16

$6 x \cdot 1 + 4 x \cdot 2 = 8$ $- 12 x \cdot 1 - 8 x \cdot 2 = - 16$

Step 1

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

Simplify each term.

Multiply $6$ by $1$ .

$6 x + 4 x \cdot 2 = 8 - 12 x \cdot 1 - 8 x \cdot 2 = - 16$

Multiply $2$ by $4$ .

$6 x + 8 x = 8 - 12 x \cdot 1 - 8 x \cdot 2 = - 16$

Add $6 x$ and $8 x$ .

$14 x = 8 - 12 x \cdot 1 - 8 x \cdot 2 = - 16$

Simplify each term.

Multiply $- 12$ by $1$ .

$14 x = 8 - 12 x - 8 x \cdot 2 = - 16$

Multiply $2$ by $- 8$ .

$14 x = 8 - 12 x - 16 x = - 16$

Subtract $16 x$ from $- 12 x$ .

$14 x = 8 - 28 x = - 16$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 14 & 8 \\ - 28 & - 16 \end{array}]$

Step 3

Find the reduced row echelon form.

Multiply each element of $R_{1}$ by $\frac{1}{14}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{14}{14} & \frac{8}{14} \\ - 28 & - 16 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{4}{7} \\ - 28 & - 16 \end{array}]$

Perform the row operation $R_{2} = R_{2} + 28 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{7} \\ - 28 + 28 \cdot 1 & - 16 + 28 (\frac{4}{7}) \end{array}]$

Simplify $R_{2}$ .

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

Step 4

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

$x = \frac{4}{7} 0 = 0$

Step 5

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

$(error = error)$

Solve Using an Augmented Matrix 6x*1+4x*2=8 -12x*1-8x*2=-16