Precalculus Examples

$y - z = 3 - 2 x$ , $3 x - y - 3 = - 3 y$ , $2 z - 3 = x + 3 y$

Step 1

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

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Step 1.1

Add $2 x$ to both sides of the equation.

$y - z + 2 x = 3$

$3 x - y - 3 = - 3 y$

$2 z - 3 = x + 3 y$

Step 1.2

Move $- z$ .

$y + 2 x - z = 3$

$3 x - y - 3 = - 3 y$

$2 z - 3 = x + 3 y$

Step 1.3

Reorder $y$ and $2 x$ .

$2 x + y - z = 3$

$3 x - y - 3 = - 3 y$

$2 z - 3 = x + 3 y$

Step 1.4

Add $3 y$ to both sides of the equation.

$2 x + y - z = 3$

$3 x - y - 3 + 3 y = 0$

$2 z - 3 = x + 3 y$

Step 1.5

Add $- y$ and $3 y$ .

$2 x + y - z = 3$

$3 x + 2 y - 3 = 0$

$2 z - 3 = x + 3 y$

Step 1.6

Add $3$ to both sides of the equation.

$2 x + y - z = 3$

$3 x + 2 y = 3$

$2 z - 3 = x + 3 y$

Step 1.7

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

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Step 1.7.1

Subtract $x$ from both sides of the equation.

$2 x + y - z = 3$

$3 x + 2 y = 3$

$2 z - 3 - x = 3 y$

Step 1.7.2

Subtract $3 y$ from both sides of the equation.

$2 x + y - z = 3$

$3 x + 2 y = 3$

$2 z - 3 - x - 3 y = 0$

$2 x + y - z = 3$

$3 x + 2 y = 3$

$2 z - 3 - x - 3 y = 0$

Step 1.8

Add $3$ to both sides of the equation.

$2 x + y - z = 3$

$3 x + 2 y = 3$

$2 z - x - 3 y = 3$

Step 1.9

Move $2 z$ .

$2 x + y - z = 3$

$3 x + 2 y = 3$

$- x - 3 y + 2 z = 3$

$2 x + y - z = 3$

$3 x + 2 y = 3$

$- x - 3 y + 2 z = 3$

Step 2

Write the system as a matrix.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 3 - 3 \cdot 1 & 2 - 3 (\frac{1}{2}) & 0 - 3 (- \frac{1}{2}) & 3 - 3 (\frac{3}{2}) \\ - 1 & - 3 & 2 & 3 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{3}{2} & - \frac{3}{2} \\ 0 & - \frac{5}{2} & \frac{3}{2} & \frac{9}{2} \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 2 \cdot 0 & 2 (\frac{1}{2}) & 2 (\frac{3}{2}) & 2 (- \frac{3}{2}) \\ 0 & - \frac{5}{2} & \frac{3}{2} & \frac{9}{2} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 3 & - 3 \\ 0 & - \frac{5}{2} & \frac{3}{2} & \frac{9}{2} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} + \frac{5}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} + \frac{5}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 3 & - 3 \\ 0 + \frac{5}{2} \cdot 0 & - \frac{5}{2} + \frac{5}{2} \cdot 1 & \frac{3}{2} + \frac{5}{2} \cdot 3 & \frac{9}{2} + \frac{5}{2} \cdot - 3 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 3 & - 3 \\ 0 & 0 & 9 & - 3 \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & 3 & - 3 \\ \frac{0}{9} & \frac{0}{9} & \frac{9}{9} & \frac{- 3}{9} \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

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

Step 3.7

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

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Step 3.7.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & - \frac{1}{2} & \frac{3}{2} \\ 0 - 3 \cdot 0 & 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 3 - 3 (- \frac{1}{3}) \\ 0 & 0 & 1 & - \frac{1}{3} \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

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

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & \frac{1}{2} + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & \frac{3}{2} + \frac{1}{2} (- \frac{1}{3}) \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & - \frac{1}{3} \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

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

Step 3.9

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

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Step 3.9.1

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

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

Step 3.9.2

Simplify $R_{1}$ .

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

Step 4

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

$x = \frac{7}{3}$

$y = - 2$

$z = - \frac{1}{3}$

Step 5

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

$(\frac{7}{3}, - 2, - \frac{1}{3})$