Finite Math Examples

$x - 8 y = - 21$ ,

$- 4 x + 2 y = - 18$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & - 8 \\ - 4 & 2 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 21 \\ - 18 \end{array}]$

Step 2

Find the determinant of the coefficient matrix

$[\begin{array}{c} 1 & - 8 \\ - 4 & 2 \end{array}]$ .

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Step 2.1

Write

$[\begin{array}{c} 1 & - 8 \\ - 4 & 2 \end{array}]$ in determinant notation.

$| \begin{array}{c} 1 & - 8 \\ - 4 & 2 \end{array} |$

Step 2.2

The determinant of a

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

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

$1 \cdot 2 - (- 4 \cdot - 8)$

Step 2.3

Simplify the determinant.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Multiply

$2$ by

$1$ .

$2 - (- 4 \cdot - 8)$

Step 2.3.1.2

Multiply

$- (- 4 \cdot - 8)$ .

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Step 2.3.1.2.1

Multiply

$- 4$ by

$- 8$ .

$2 - 1 \cdot 32$

Step 2.3.1.2.2

Multiply

$- 1$ by

$32$ .

$2 - 32$

Step 2.3.2

Subtract

$32$ from

$2$ .

$- 30$

$D = - 30$

Step 3

Since the determinant is not

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

Step 4

Find the value of

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

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

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Step 4.1

Replace column

$1$ of the coefficient matrix that corresponds to the

$x$ -coefficients of the system with

$[\begin{array}{c} - 21 \\ - 18 \end{array}]$ .

$| \begin{array}{c} - 21 & - 8 \\ - 18 & 2 \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

The determinant of a

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

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

$- 21 \cdot 2 - (- 18 \cdot - 8)$

Step 4.2.2

Simplify the determinant.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply

$- 21$ by

$2$ .

$- 42 - (- 18 \cdot - 8)$

Step 4.2.2.1.2

Multiply

$- (- 18 \cdot - 8)$ .

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Step 4.2.2.1.2.1

Multiply

$- 18$ by

$- 8$ .

$- 42 - 1 \cdot 144$

Step 4.2.2.1.2.2

Multiply

$- 1$ by

$144$ .

$- 42 - 144$

Step 4.2.2.2

Subtract

$144$ from

$- 42$ .

$- 186$

$D_{x} = - 186$

Step 4.3

Use the formula to solve for

$x$ .

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

Step 4.4

Substitute

$- 30$ for

$D$ and

$- 186$ for

$D_{x}$ in the formula.

$x = \frac{- 186}{- 30}$

Step 4.5

Cancel the common factor of

$- 186$ and

$- 30$ .

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Step 4.5.1

Factor

$- 6$ out of

$- 186$ .

$x = \frac{- 6 (31)}{- 30}$

Step 4.5.2

Cancel the common factors.

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Step 4.5.2.1

Factor

$- 6$ out of

$- 30$ .

$x = \frac{- 6 \cdot 31}{- 6 \cdot 5}$

Step 4.5.2.2

Cancel the common factor.

$x = \frac{- 6 \cdot 31}{- 6 \cdot 5}$

Step 4.5.2.3

Rewrite the expression.

$x = \frac{31}{5}$

Step 5

Find the value of

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

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

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Step 5.1

Replace column

$2$ of the coefficient matrix that corresponds to the

$y$ -coefficients of the system with

$[\begin{array}{c} - 21 \\ - 18 \end{array}]$ .

$| \begin{array}{c} 1 & - 21 \\ - 4 & - 18 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

The determinant of a

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

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

$1 \cdot - 18 - (- 4 \cdot - 21)$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply

$- 18$ by

$1$ .

$- 18 - (- 4 \cdot - 21)$

Step 5.2.2.1.2

Multiply

$- (- 4 \cdot - 21)$ .

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Step 5.2.2.1.2.1

Multiply

$- 4$ by

$- 21$ .

$- 18 - 1 \cdot 84$

Step 5.2.2.1.2.2

Multiply

$- 1$ by

$84$ .

$- 18 - 84$

Step 5.2.2.2

Subtract

$84$ from

$- 18$ .

$- 102$

$D_{y} = - 102$

Step 5.3

Use the formula to solve for

$y$ .

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

Step 5.4

Substitute

$- 30$ for

$D$ and

$- 102$ for

$D_{y}$ in the formula.

$y = \frac{- 102}{- 30}$

Step 5.5

Cancel the common factor of

$- 102$ and

$- 30$ .

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Step 5.5.1

Factor

$- 6$ out of

$- 102$ .

$y = \frac{- 6 (17)}{- 30}$

Step 5.5.2

Cancel the common factors.

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Step 5.5.2.1

Factor

$- 6$ out of

$- 30$ .

$y = \frac{- 6 \cdot 17}{- 6 \cdot 5}$

Step 5.5.2.2

Cancel the common factor.

$y = \frac{- 6 \cdot 17}{- 6 \cdot 5}$

Step 5.5.2.3

Rewrite the expression.

$y = \frac{17}{5}$

Step 6

List the solution to the system of equations.

$x = \frac{31}{5}$

$y = \frac{17}{5}$