Precalculus Examples

$x + y + z = 8$ , $x - y + z = - 2$ , $2 x + 0 + 2 = 9$

Step 1

Add $2 x$ and $0$ .

$x + y + z = 8, x - y + z = - 2, 2 x + 2 = 9$

Step 2

Subtract $2$ from both sides of the equation.

$x + y + z = 8, x - y + z = - 2, 2 x = 9 - 2$

Step 3

Subtract $2$ from $9$ .

$x + y + z = 8, x - y + z = - 2, 2 x = 7$

Step 4

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 2 & 0 & 0 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 8 \\ - 2 \\ 7 \end{array}]$

Step 5

Find the determinant of $[\begin{array}{c} 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 2 & 0 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$D = - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Simplify each term.

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The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D = - 1 (1 \cdot 0 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $0$ by $1$ .

$D = - 1 (0 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Multiply $- 2$ by $1$ .

$D = - 1 (0 - 2) - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Subtract $2$ from $0$ .

$D = - 1 \cdot - 2 - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Multiply $- 1$ by $- 2$ .

$D = 2 - 1 | \begin{array}{c} 1 & 1 \\ 2 & 0 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D = 2 - 1 (1 \cdot 0 - 2 \cdot 1) + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $0$ by $1$ .

$D = 2 - 1 (0 - 2 \cdot 1) + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Multiply $- 2$ by $1$ .

$D = 2 - 1 (0 - 2) + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Subtract $2$ from $0$ .

$D = 2 - 1 \cdot - 2 + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

Multiply $- 1$ by $- 2$ .

$D = 2 + 2 + 0 | \begin{array}{c} 1 & 1 \\ 1 & 1 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D = 2 + 2 + 0 (1 \cdot 1 - 1 \cdot 1)$

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ .

$D = 2 + 2 + 0 (1 - 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$D = 2 + 2 + 0 (1 - 1)$

Subtract $1$ from $1$ .

$D = 2 + 2 + 0 \cdot 0$

Multiply $0$ by $0$ .

$D = 2 + 2 + 0$

Add $2$ and $2$ .

$D = 4 + 0$

Add $4$ and $0$ .

$D = 4$

Step 6

Find the determinant of $[\begin{array}{c} 8 & 1 & 1 \\ - 2 & - 1 & 1 \\ 7 & 0 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$D_{x} = - 1 | \begin{array}{c} - 2 & 1 \\ 7 & 0 \end{array} | - 1 | \begin{array}{c} 8 & 1 \\ 7 & 0 \end{array} | + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Simplify each term.

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The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{x} = - 1 (- 2 \cdot 0 - 7 \cdot 1) - 1 | \begin{array}{c} 8 & 1 \\ 7 & 0 \end{array} | + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 2$ by $0$ .

$D_{x} = - 1 (0 - 7 \cdot 1) - 1 | \begin{array}{c} 8 & 1 \\ 7 & 0 \end{array} | + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Multiply $- 7$ by $1$ .

$D_{x} = - 1 (0 - 7) - 1 | \begin{array}{c} 8 & 1 \\ 7 & 0 \end{array} | + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Subtract $7$ from $0$ .

$D_{x} = - 1 \cdot - 7 - 1 | \begin{array}{c} 8 & 1 \\ 7 & 0 \end{array} | + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Multiply $- 1$ by $- 7$ .

$D_{x} = 7 - 1 | \begin{array}{c} 8 & 1 \\ 7 & 0 \end{array} | + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{x} = 7 - 1 (8 \cdot 0 - 7 \cdot 1) + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $8$ by $0$ .

$D_{x} = 7 - 1 (0 - 7 \cdot 1) + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Multiply $- 7$ by $1$ .

$D_{x} = 7 - 1 (0 - 7) + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Subtract $7$ from $0$ .

$D_{x} = 7 - 1 \cdot - 7 + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

Multiply $- 1$ by $- 7$ .

$D_{x} = 7 + 7 + 0 | \begin{array}{c} 8 & 1 \\ - 2 & 1 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{x} = 7 + 7 + 0 (8 \cdot 1 - (- 2 \cdot 1))$

Simplify the determinant.

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Simplify each term.

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Multiply $8$ by $1$ .

$D_{x} = 7 + 7 + 0 (8 - (- 2 \cdot 1))$

Multiply $- (- 2 \cdot 1)$ .

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Multiply $- 2$ by $1$ .

$D_{x} = 7 + 7 + 0 (8 + 2)$

Multiply $- 1$ by $- 2$ .

$D_{x} = 7 + 7 + 0 (8 + 2)$

Add $8$ and $2$ .

$D_{x} = 7 + 7 + 0 \cdot 10$

Multiply $0$ by $10$ .

$D_{x} = 7 + 7 + 0$

Add $7$ and $7$ .

$D_{x} = 14 + 0$

Add $14$ and $0$ .

$D_{x} = 14$

Step 7

Find the determinant of $[\begin{array}{c} 1 & 8 & 1 \\ 1 & - 2 & 1 \\ 2 & 7 & 0 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$D_{y} = 1 | \begin{array}{c} 1 & - 2 \\ 2 & 7 \end{array} | - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Simplify each term.

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Multiply $| \begin{array}{c} 1 & - 2 \\ 2 & 7 \end{array} |$ by $1$ .

$D_{y} = | \begin{array}{c} 1 & - 2 \\ 2 & 7 \end{array} | - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{y} = 1 \cdot 7 - 2 \cdot - 2 - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $7$ by $1$ .

$D_{y} = 7 - 2 \cdot - 2 - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 2$ by $- 2$ .

$D_{y} = 7 + 4 - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Add $7$ and $4$ .

$D_{y} = 11 - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{y} = 11 - 1 (1 \cdot 7 - 2 \cdot 8) + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $7$ by $1$ .

$D_{y} = 11 - 1 (7 - 2 \cdot 8) + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 2$ by $8$ .

$D_{y} = 11 - 1 (7 - 16) + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Subtract $16$ from $7$ .

$D_{y} = 11 - 1 \cdot - 9 + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 1$ by $- 9$ .

$D_{y} = 11 + 9 + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{y} = 11 + 9 + 0 (1 \cdot - 2 - 1 \cdot 8)$

Simplify the determinant.

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Simplify each term.

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Multiply $- 2$ by $1$ .

$D_{y} = 11 + 9 + 0 (- 2 - 1 \cdot 8)$

Multiply $- 1$ by $8$ .

$D_{y} = 11 + 9 + 0 (- 2 - 8)$

Subtract $8$ from $- 2$ .

$D_{y} = 11 + 9 + 0 \cdot - 10$

Multiply $0$ by $- 10$ .

$D_{y} = 11 + 9 + 0$

Add $11$ and $9$ .

$D_{y} = 20 + 0$

Add $20$ and $0$ .

$D_{y} = 20$

Step 8

Find the determinant of $[\begin{array}{c} 1 & 1 & 8 \\ 1 & - 1 & - 2 \\ 2 & 0 & 7 \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

$D_{z} = - 1 | \begin{array}{c} 1 & - 2 \\ 2 & 7 \end{array} | - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Simplify each term.

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The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{z} = - 1 (1 \cdot 7 - 2 \cdot - 2) - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $7$ by $1$ .

$D_{z} = - 1 (7 - 2 \cdot - 2) - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 2$ by $- 2$ .

$D_{z} = - 1 (7 + 4) - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Add $7$ and $4$ .

$D_{z} = - 1 \cdot 11 - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 1$ by $11$ .

$D_{z} = - 11 - 1 | \begin{array}{c} 1 & 8 \\ 2 & 7 \end{array} | + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{z} = - 11 - 1 (1 \cdot 7 - 2 \cdot 8) + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $7$ by $1$ .

$D_{z} = - 11 - 1 (7 - 2 \cdot 8) + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 2$ by $8$ .

$D_{z} = - 11 - 1 (7 - 16) + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Subtract $16$ from $7$ .

$D_{z} = - 11 - 1 \cdot - 9 + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

Multiply $- 1$ by $- 9$ .

$D_{z} = - 11 + 9 + 0 | \begin{array}{c} 1 & 8 \\ 1 & - 2 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{z} = - 11 + 9 + 0 (1 \cdot - 2 - 1 \cdot 8)$

Simplify the determinant.

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Simplify each term.

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Multiply $- 2$ by $1$ .

$D_{z} = - 11 + 9 + 0 (- 2 - 1 \cdot 8)$

Multiply $- 1$ by $8$ .

$D_{z} = - 11 + 9 + 0 (- 2 - 8)$

Subtract $8$ from $- 2$ .

$D_{z} = - 11 + 9 + 0 \cdot - 10$

Multiply $0$ by $- 10$ .

$D_{z} = - 11 + 9 + 0$

Add $- 11$ and $9$ .

$D_{z} = - 2 + 0$

Add $- 2$ and $0$ .

$D_{z} = - 2$

Step 9

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{14}{4}$ .

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Remove parentheses.

$x = \frac{14}{4}$

Cancel the common factor of $14$ and $4$ .

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Factor $2$ out of $14$ .

$x = \frac{2 (7)}{4}$

Cancel the common factors.

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Factor $2$ out of $4$ .

$x = \frac{2 \cdot 7}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot 7}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{7}{2}$

Step 10

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{20}{4}$ .

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Remove parentheses.

$y = \frac{20}{4}$

Divide $20$ by $4$ .

$y = 5$

Step 11

Find the value of $z$ by Cramer's Rule, which states that $z = \frac{D_{z}}{D}$ . In this case, $z = \frac{- 2}{4}$ .

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Remove parentheses.

$z = \frac{- 2}{4}$

Simplify $\frac{- 2}{4}$ .

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Cancel the common factor of $- 2$ and $4$ .

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Factor $2$ out of $- 2$ .

$z = \frac{2 (- 1)}{4}$

Cancel the common factors.

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Factor $2$ out of $4$ .

$z = \frac{2 \cdot - 1}{2 \cdot 2}$

Cancel the common factor.

$z = \frac{2 \cdot - 1}{2 \cdot 2}$

Rewrite the expression.

$z = \frac{- 1}{2}$

Move the negative in front of the fraction.

$z = - \frac{1}{2}$

Step 12

The solution to the system of equations using Cramer's Rule.

$x = \frac{7}{2}, y = 5, z = - \frac{1}{2}$