Precalculus Examples

$x - 8 = 8$ , $4 x + 3 y = 2$

Step 1

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

Tap for more steps...

Step 1.1

Add $8$ to both sides of the equation.

$x = 8 + 8$

$4 x + 3 y = 2$

Step 1.2

Add $8$ and $8$ .

$x = 16$

$4 x + 3 y = 2$

$x = 16$

$4 x + 3 y = 2$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 0 & 16 \\ 4 & 3 & 2 \end{array}]$

Step 3

Find the reduced row echelon form.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

$[\begin{array}{c} 1 & 0 & 16 \\ 4 - 4 \cdot 1 & 3 - 4 \cdot 0 & 2 - 4 \cdot 16 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 16 \\ 0 & 3 & - 62 \end{array}]$

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

$[\begin{array}{c} 1 & 0 & 16 \\ \frac{0}{3} & \frac{3}{3} & \frac{- 62}{3} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 16 \\ 0 & 1 & - \frac{62}{3} \end{array}]$

Step 4

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

$x = 16$

$y = - \frac{62}{3}$

Step 5

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

$(16, - \frac{62}{3})$