Precalculus Examples

$x + 7 y = z + 6$ , $x = 2 + y - z$ , $x + y - 3 z = - 4$

Step 1

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

Tap for more steps...

Step 1.1

Subtract $z$ from both sides of the equation.

$x + 7 y - z = 6$

$x = 2 + y - z$

$x + y - 3 z = - 4$

Step 1.2

Move all terms containing variables to the left side of the equation.

Tap for more steps...

Step 1.2.1

Subtract $y$ from both sides of the equation.

$x + 7 y - z = 6$

$x - y = 2 - z$

$x + y - 3 z = - 4$

Step 1.2.2

Add $z$ to both sides of the equation.

$x + 7 y - z = 6$

$x - y + z = 2$

$x + y - 3 z = - 4$

$x + 7 y - z = 6$

$x - y + z = 2$

$x + y - 3 z = - 4$

$x + 7 y - z = 6$

$x - y + z = 2$

$x + y - 3 z = - 4$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 1 & - 1 & 1 & 2 \\ 1 & 1 & - 3 & - 4 \end{array}]$

Step 3

Find the reduced row echelon form.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 1 - 1 & - 1 - 7 & 1 + 1 & 2 - 6 \\ 1 & 1 & - 3 & - 4 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & - 8 & 2 & - 4 \\ 1 & 1 & - 3 & - 4 \end{array}]$

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & - 8 & 2 & - 4 \\ 1 - 1 & 1 - 7 & - 3 + 1 & - 4 - 6 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & - 8 & 2 & - 4 \\ 0 & - 6 & - 2 & - 10 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ - \frac{1}{8} \cdot 0 & - \frac{1}{8} \cdot - 8 & - \frac{1}{8} \cdot 2 & - \frac{1}{8} \cdot - 4 \\ 0 & - 6 & - 2 & - 10 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & 1 & - \frac{1}{4} & \frac{1}{2} \\ 0 & - 6 & - 2 & - 10 \end{array}]$

Step 3.4

Perform the row operation $R_{3} = R_{3} + 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

Tap for more steps...

Step 3.4.1

Perform the row operation $R_{3} = R_{3} + 6 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & 1 & - \frac{1}{4} & \frac{1}{2} \\ 0 + 6 \cdot 0 & - 6 + 6 \cdot 1 & - 2 + 6 (- \frac{1}{4}) & - 10 + 6 (\frac{1}{2}) \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & 1 & - \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & - \frac{7}{2} & - 7 \end{array}]$

Step 3.5

Multiply each element of $R_{3}$ by $- \frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

Tap for more steps...

Step 3.5.1

Multiply each element of $R_{3}$ by $- \frac{2}{7}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & 1 & - \frac{1}{4} & \frac{1}{2} \\ - \frac{2}{7} \cdot 0 & - \frac{2}{7} \cdot 0 & - \frac{2}{7} (- \frac{7}{2}) & - \frac{2}{7} \cdot - 7 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & 1 & - \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3.6

Perform the row operation $R_{2} = R_{2} + \frac{1}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

Tap for more steps...

Step 3.6.1

Perform the row operation $R_{2} = R_{2} + \frac{1}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 + \frac{1}{4} \cdot 0 & 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & \frac{1}{2} + \frac{1}{4} \cdot 2 \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 7 & - 1 & 6 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3.7

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

Tap for more steps...

Step 3.7.1

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

$[\begin{array}{c} 1 + 0 & 7 + 0 & - 1 + 1 \cdot 1 & 6 + 1 \cdot 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 7 & 0 & 8 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3.8

Perform the row operation $R_{1} = R_{1} - 7 R_{2}$ to make the entry at $1, 2$ a $0$ .

Tap for more steps...

Step 3.8.1

Perform the row operation $R_{1} = R_{1} - 7 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 7 \cdot 0 & 7 - 7 \cdot 1 & 0 - 7 \cdot 0 & 8 - 7 \cdot 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 1$

$y = 1$

$z = 2$

Step 5

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

$(1, 1, 2)$