Finite Math Examples

9 y - 5 x = 3

$9 y - 5 x = 3$ ,

x + y = 1

$x + y = 1$ ,

z + 2 y = 2

$z + 2 y = 2$

Step 1

Move all of the variables to the left side of each equation.

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Step 1.1

Reorder

9 y

$9 y$ and

- 5 x

$- 5 x$ .

- 5 x + 9 y = 3

$- 5 x + 9 y = 3$

x + y = 1

$x + y = 1$

z + 2 y = 2

$z + 2 y = 2$

Step 1.2

Reorder

z

$z$ and

2 y

$2 y$ .

- 5 x + 9 y = 3

$- 5 x + 9 y = 3$

x + y = 1

$x + y = 1$

2 y + z = 2

$2 y + z = 2$

- 5 x + 9 y = 3

$- 5 x + 9 y = 3$

x + y = 1

$x + y = 1$

2 y + z = 2

$2 y + z = 2$

Step 2

Represent the system of equations in matrix format.

⎡ ⎢ ⎣ \begin{matrix} - 5 & 9 & 0 1 & 1 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 312 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} - 5 & 9 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 3 \\ 1 \\ 2 \end{array}]$

Step 3

Find the determinant of the coefficient matrix

⎡ ⎢ ⎣ \begin{matrix} - 5 & 9 & 0 1 & 1 & 0 0 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

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

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Step 3.1

Write

$[\begin{array}{c} - 5 & 9 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} - 5 & 9 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{array} |$

Step 3.2

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in column

$3$ by its cofactor and add.

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Step 3.2.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 3.2.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 3.2.3

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

Step 3.2.4

Multiply element

$a_{13}$ by its cofactor.

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

Step 3.2.5

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} - 5 & 9 \\ 0 & 2 \end{array} |$

Step 3.2.6

Multiply element

$a_{23}$ by its cofactor.

$0 | \begin{array}{c} - 5 & 9 \\ 0 & 2 \end{array} |$

Step 3.2.7

The minor for

$a_{33}$ is the determinant with row

$3$ and column

$3$ deleted.

$| \begin{array}{c} - 5 & 9 \\ 1 & 1 \end{array} |$

Step 3.2.8

Multiply element

$a_{33}$ by its cofactor.

$1 | \begin{array}{c} - 5 & 9 \\ 1 & 1 \end{array} |$

Step 3.2.9

Add the terms together.

$0 | \begin{array}{c} 1 & 1 \\ 0 & 2 \end{array} | + 0 | \begin{array}{c} - 5 & 9 \\ 0 & 2 \end{array} | + 1 | \begin{array}{c} - 5 & 9 \\ 1 & 1 \end{array} |$

Step 3.3

Multiply

$0$ by

$| \begin{array}{c} 1 & 1 \\ 0 & 2 \end{array} |$ .

$0 + 0 | \begin{array}{c} - 5 & 9 \\ 0 & 2 \end{array} | + 1 | \begin{array}{c} - 5 & 9 \\ 1 & 1 \end{array} |$

Step 3.4

Multiply

$0$ by

$| \begin{array}{c} - 5 & 9 \\ 0 & 2 \end{array} |$ .

$0 + 0 + 1 | \begin{array}{c} - 5 & 9 \\ 1 & 1 \end{array} |$

Step 3.5

Evaluate

$| \begin{array}{c} - 5 & 9 \\ 1 & 1 \end{array} |$ .

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Step 3.5.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$0 + 0 + 1 (- 5 \cdot 1 - 1 \cdot 9)$

Step 3.5.2

Simplify the determinant.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

Multiply

$- 5$ by

$1$ .

$0 + 0 + 1 (- 5 - 1 \cdot 9)$

Step 3.5.2.1.2

Multiply

$- 1$ by

$9$ .

$0 + 0 + 1 (- 5 - 9)$

Step 3.5.2.2

Subtract

$9$ from

$- 5$ .

$0 + 0 + 1 \cdot - 14$

Step 3.6

Simplify the determinant.

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Step 3.6.1

Multiply

$- 14$ by

$1$ .

$0 + 0 - 14$

Step 3.6.2

Add

$0$ and

$0$ .

$0 - 14$

Step 3.6.3

Subtract

$14$ from

$0$ .

$- 14$

$D = - 14$

Step 4

Since the determinant is not

$0$ , the system can be solved using Cramer's Rule.

Step 5

Find the value of

$x$ by Cramer's Rule, which states that

$x = \frac{D_{x}}{D}$ .

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Step 5.1

Replace column

$1$ of the coefficient matrix that corresponds to the

$x$ -coefficients of the system with

$[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}]$ .

$| \begin{array}{c} 3 & 9 & 0 \\ 1 & 1 & 0 \\ 2 & 2 & 1 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in column

$3$ by its cofactor and add.

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Step 5.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 5.2.1.3

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} |$

Step 5.2.1.4

Multiply element

$a_{13}$ by its cofactor.

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

Step 5.2.1.5

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} 3 & 9 \\ 2 & 2 \end{array} |$

Step 5.2.1.6

Multiply element

$a_{23}$ by its cofactor.

$0 | \begin{array}{c} 3 & 9 \\ 2 & 2 \end{array} |$

Step 5.2.1.7

The minor for

$a_{33}$ is the determinant with row

$3$ and column

$3$ deleted.

$| \begin{array}{c} 3 & 9 \\ 1 & 1 \end{array} |$

Step 5.2.1.8

Multiply element

$a_{33}$ by its cofactor.

$1 | \begin{array}{c} 3 & 9 \\ 1 & 1 \end{array} |$

Step 5.2.1.9

Add the terms together.

$0 | \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} | + 0 | \begin{array}{c} 3 & 9 \\ 2 & 2 \end{array} | + 1 | \begin{array}{c} 3 & 9 \\ 1 & 1 \end{array} |$

Step 5.2.2

Multiply

$0$ by

$| \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} |$ .

$0 + 0 | \begin{array}{c} 3 & 9 \\ 2 & 2 \end{array} | + 1 | \begin{array}{c} 3 & 9 \\ 1 & 1 \end{array} |$

Step 5.2.3

Multiply

$0$ by

$| \begin{array}{c} 3 & 9 \\ 2 & 2 \end{array} |$ .

$0 + 0 + 1 | \begin{array}{c} 3 & 9 \\ 1 & 1 \end{array} |$

Step 5.2.4

Evaluate

$| \begin{array}{c} 3 & 9 \\ 1 & 1 \end{array} |$ .

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Step 5.2.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$0 + 0 + 1 (3 \cdot 1 - 1 \cdot 9)$

Step 5.2.4.2

Simplify the determinant.

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Step 5.2.4.2.1

Simplify each term.

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Step 5.2.4.2.1.1

Multiply

$3$ by

$1$ .

$0 + 0 + 1 (3 - 1 \cdot 9)$

Step 5.2.4.2.1.2

Multiply

$- 1$ by

$9$ .

$0 + 0 + 1 (3 - 9)$

Step 5.2.4.2.2

Subtract

$9$ from

$3$ .

$0 + 0 + 1 \cdot - 6$

Step 5.2.5

Simplify the determinant.

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Step 5.2.5.1

Multiply

$- 6$ by

$1$ .

$0 + 0 - 6$

Step 5.2.5.2

Add

$0$ and

$0$ .

$0 - 6$

Step 5.2.5.3

Subtract

$6$ from

$0$ .

$- 6$

$D_{x} = - 6$

Step 5.3

Use the formula to solve for

$x$ .

$x = \frac{D_{x}}{D}$

Step 5.4

Substitute

$- 14$ for

$D$ and

$- 6$ for

$D_{x}$ in the formula.

$x = \frac{- 6}{- 14}$

Step 5.5

Cancel the common factor of

$- 6$ and

$- 14$ .

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Step 5.5.1

Factor

$- 2$ out of

$- 6$ .

$x = \frac{- 2 (3)}{- 14}$

Step 5.5.2

Cancel the common factors.

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Step 5.5.2.1

Factor

$- 2$ out of

$- 14$ .

$x = \frac{- 2 \cdot 3}{- 2 \cdot 7}$

Step 5.5.2.2

Cancel the common factor.

$x = \frac{- 2 \cdot 3}{- 2 \cdot 7}$

Step 5.5.2.3

Rewrite the expression.

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

Step 6

Find the value of

$y$ by Cramer's Rule, which states that

$y = \frac{D_{y}}{D}$ .

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Step 6.1

Replace column

$2$ of the coefficient matrix that corresponds to the

$y$ -coefficients of the system with

$[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}]$ .

$| \begin{array}{c} - 5 & 3 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in column

$3$ by its cofactor and add.

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Step 6.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 6.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 6.2.1.3

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

Step 6.2.1.4

Multiply element

$a_{13}$ by its cofactor.

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

Step 6.2.1.5

The minor for

$a_{23}$ is the determinant with row

$2$ and column

$3$ deleted.

$| \begin{array}{c} - 5 & 3 \\ 0 & 2 \end{array} |$

Step 6.2.1.6

Multiply element

$a_{23}$ by its cofactor.

$0 | \begin{array}{c} - 5 & 3 \\ 0 & 2 \end{array} |$

Step 6.2.1.7

The minor for

$a_{33}$ is the determinant with row

$3$ and column

$3$ deleted.

$| \begin{array}{c} - 5 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.1.8

Multiply element

$a_{33}$ by its cofactor.

$1 | \begin{array}{c} - 5 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.1.9

Add the terms together.

$0 | \begin{array}{c} 1 & 1 \\ 0 & 2 \end{array} | + 0 | \begin{array}{c} - 5 & 3 \\ 0 & 2 \end{array} | + 1 | \begin{array}{c} - 5 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.2

Multiply

$0$ by

$| \begin{array}{c} 1 & 1 \\ 0 & 2 \end{array} |$ .

$0 + 0 | \begin{array}{c} - 5 & 3 \\ 0 & 2 \end{array} | + 1 | \begin{array}{c} - 5 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.3

Multiply

$0$ by

$| \begin{array}{c} - 5 & 3 \\ 0 & 2 \end{array} |$ .

$0 + 0 + 1 | \begin{array}{c} - 5 & 3 \\ 1 & 1 \end{array} |$

Step 6.2.4

Evaluate

$| \begin{array}{c} - 5 & 3 \\ 1 & 1 \end{array} |$ .

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Step 6.2.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$0 + 0 + 1 (- 5 \cdot 1 - 1 \cdot 3)$

Step 6.2.4.2

Simplify the determinant.

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Step 6.2.4.2.1

Simplify each term.

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Step 6.2.4.2.1.1

Multiply

$- 5$ by

$1$ .

$0 + 0 + 1 (- 5 - 1 \cdot 3)$

Step 6.2.4.2.1.2

Multiply

$- 1$ by

$3$ .

$0 + 0 + 1 (- 5 - 3)$

Step 6.2.4.2.2

Subtract

$3$ from

$- 5$ .

$0 + 0 + 1 \cdot - 8$

Step 6.2.5

Simplify the determinant.

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Step 6.2.5.1

Multiply

$- 8$ by

$1$ .

$0 + 0 - 8$

Step 6.2.5.2

Add

$0$ and

$0$ .

$0 - 8$

Step 6.2.5.3

Subtract

$8$ from

$0$ .

$- 8$

$D_{y} = - 8$

Step 6.3

Use the formula to solve for

$y$ .

$y = \frac{D_{y}}{D}$

Step 6.4

Substitute

$- 14$ for

$D$ and

$- 8$ for

$D_{y}$ in the formula.

$y = \frac{- 8}{- 14}$

Step 6.5

Cancel the common factor of

$- 8$ and

$- 14$ .

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Step 6.5.1

Factor

$- 2$ out of

$- 8$ .

$y = \frac{- 2 (4)}{- 14}$

Step 6.5.2

Cancel the common factors.

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Step 6.5.2.1

Factor

$- 2$ out of

$- 14$ .

$y = \frac{- 2 \cdot 4}{- 2 \cdot 7}$

Step 6.5.2.2

Cancel the common factor.

$y = \frac{- 2 \cdot 4}{- 2 \cdot 7}$

Step 6.5.2.3

Rewrite the expression.

$y = \frac{4}{7}$

Step 7

Find the value of

$z$ by Cramer's Rule, which states that

$z = \frac{D_{z}}{D}$ .

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Step 7.1

Replace column

$3$ of the coefficient matrix that corresponds to the

$z$ -coefficients of the system with

$[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}]$ .

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

Step 7.2

Find the determinant.

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Step 7.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in column

$1$ by its cofactor and add.

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Step 7.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 7.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 7.2.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} |$

Step 7.2.1.4

Multiply element

$a_{11}$ by its cofactor.

$- 5 | \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} |$

Step 7.2.1.5

The minor for

$a_{21}$ is the determinant with row

$2$ and column

$1$ deleted.

$| \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} |$

Step 7.2.1.6

Multiply element

$a_{21}$ by its cofactor.

$- 1 | \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} |$

Step 7.2.1.7

The minor for

$a_{31}$ is the determinant with row

$3$ and column

$1$ deleted.

$| \begin{array}{c} 9 & 3 \\ 1 & 1 \end{array} |$

Step 7.2.1.8

Multiply element

$a_{31}$ by its cofactor.

$0 | \begin{array}{c} 9 & 3 \\ 1 & 1 \end{array} |$

Step 7.2.1.9

Add the terms together.

$- 5 | \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} | - 1 | \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} | + 0 | \begin{array}{c} 9 & 3 \\ 1 & 1 \end{array} |$

Step 7.2.2

Multiply

$0$ by

$| \begin{array}{c} 9 & 3 \\ 1 & 1 \end{array} |$ .

$- 5 | \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} | - 1 | \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} | + 0$

Step 7.2.3

Evaluate

$| \begin{array}{c} 1 & 1 \\ 2 & 2 \end{array} |$ .

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Step 7.2.3.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 (1 \cdot 2 - 2 \cdot 1) - 1 | \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} | + 0$

Step 7.2.3.2

Simplify the determinant.

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Step 7.2.3.2.1

Simplify each term.

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Step 7.2.3.2.1.1

Multiply

$2$ by

$1$ .

$- 5 (2 - 2 \cdot 1) - 1 | \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} | + 0$

Step 7.2.3.2.1.2

Multiply

$- 2$ by

$1$ .

$- 5 (2 - 2) - 1 | \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} | + 0$

Step 7.2.3.2.2

Subtract

$2$ from

$2$ .

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

Step 7.2.4

Evaluate

$| \begin{array}{c} 9 & 3 \\ 2 & 2 \end{array} |$ .

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Step 7.2.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 \cdot 0 - 1 (9 \cdot 2 - 2 \cdot 3) + 0$

Step 7.2.4.2

Simplify the determinant.

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Step 7.2.4.2.1

Simplify each term.

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Step 7.2.4.2.1.1

Multiply

$9$ by

$2$ .

$- 5 \cdot 0 - 1 (18 - 2 \cdot 3) + 0$

Step 7.2.4.2.1.2

Multiply

$- 2$ by

$3$ .

$- 5 \cdot 0 - 1 (18 - 6) + 0$

Step 7.2.4.2.2

Subtract

$6$ from

$18$ .

$- 5 \cdot 0 - 1 \cdot 12 + 0$

Step 7.2.5

Simplify the determinant.

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Step 7.2.5.1

Simplify each term.

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Step 7.2.5.1.1

Multiply

$- 5$ by

$0$ .

$0 - 1 \cdot 12 + 0$

Step 7.2.5.1.2

Multiply

$- 1$ by

$12$ .

$0 - 12 + 0$

Step 7.2.5.2

Subtract

$12$ from

$0$ .

$- 12 + 0$

Step 7.2.5.3

Add

$- 12$ and

$0$ .

$- 12$

$D_{z} = - 12$

Step 7.3

Use the formula to solve for

$z$ .

$z = \frac{D_{z}}{D}$

Step 7.4

Substitute

$- 14$ for

$D$ and

$- 12$ for

$D_{z}$ in the formula.

$z = \frac{- 12}{- 14}$

Step 7.5

Cancel the common factor of

$- 12$ and

$- 14$ .

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Step 7.5.1

Factor

$- 2$ out of

$- 12$ .

$z = \frac{- 2 (6)}{- 14}$

Step 7.5.2

Cancel the common factors.

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Step 7.5.2.1

Factor

$- 2$ out of

$- 14$ .

$z = \frac{- 2 \cdot 6}{- 2 \cdot 7}$

Step 7.5.2.2

Cancel the common factor.

$z = \frac{- 2 \cdot 6}{- 2 \cdot 7}$

Step 7.5.2.3

Rewrite the expression.

$z = \frac{6}{7}$

Step 8

List the solution to the system of equations.

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

$y = \frac{4}{7}$

$z = \frac{6}{7}$