Finite Math Examples

5 x - y = - 2

$5 x - y = - 2$ ,

3 x - 4 y = 0

$3 x - 4 y = 0$

Step 1

Represent the system of equations in matrix format.

[\begin{matrix} 5 & - 1 3 & - 4 \end{matrix}] [\begin{matrix} x y \end{matrix}] = [\begin{matrix} - 2 0 \end{matrix}]

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]$

Step 2

Find the determinant of the coefficient matrix

[\begin{matrix} 5 & - 1 3 & - 4 \end{matrix}]

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}]$ .

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Step 2.1

Write

[\begin{matrix} 5 & - 1 3 & - 4 \end{matrix}]

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}]$ in determinant notation.

∣ ∣ ∣ \begin{matrix} 5 & - 1 3 & - 4 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array} |$

Step 2.2

The determinant of a

2 \times 2

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

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

5 \cdot - 4 - 3 \cdot - 1

$5 \cdot - 4 - 3 \cdot - 1$

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

5

$5$ by

- 4

$- 4$ .

- 20 - 3 \cdot - 1

$- 20 - 3 \cdot - 1$

Step 2.3.1.2

Multiply

- 3

$- 3$ by

- 1

$- 1$ .

- 20 + 3

$- 20 + 3$

- 20 + 3

$- 20 + 3$

Step 2.3.2

Add

- 20

$- 20$ and

3

$3$ .

- 17

$- 17$

- 17

$- 17$

D = - 17

$D = - 17$

Step 3

Since the determinant is not

0

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

Step 4

Find the value of

x

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

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

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

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Step 4.1

Replace column

1

$1$ of the coefficient matrix that corresponds to the

x

$x$ -coefficients of the system with

[\begin{matrix} - 2 0 \end{matrix}]

$[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

∣ ∣ ∣ \begin{matrix} - 2 & - 1 0 & - 4 \end{matrix} ∣ ∣ ∣

$| \begin{array}{c} - 2 & - 1 \\ 0 & - 4 \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

The determinant of a

2 \times 2

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

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

- 2 \cdot - 4 + 0 \cdot - 1

$- 2 \cdot - 4 + 0 \cdot - 1$

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

- 2

$- 2$ by

- 4

$- 4$ .

8 + 0 \cdot - 1

$8 + 0 \cdot - 1$

Step 4.2.2.1.2

Multiply

0

$0$ by

- 1

$- 1$ .

8 + 0

$8 + 0$

8 + 0

$8 + 0$

Step 4.2.2.2

Add

8

$8$ and

0

$0$ .

8

$8$

8

$8$

D_{x} = 8

$D_{x} = 8$

Step 4.3

Use the formula to solve for

x

$x$ .

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

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

Step 4.4

Substitute

- 17

$- 17$ for

D

$D$ and

8

$8$ for

D_{x}

$D_{x}$ in the formula.

x = \frac{8}{- 17}

$x = \frac{8}{- 17}$

Step 4.5

Move the negative in front of the fraction.

x = - \frac{8}{17}

$x = - \frac{8}{17}$

x = - \frac{8}{17}

$x = - \frac{8}{17}$

Step 5

Find the value of

y

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

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

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

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Step 5.1

Replace column

2

$2$ of the coefficient matrix that corresponds to the

y

$y$ -coefficients of the system with

[\begin{matrix} - 2 0 \end{matrix}]

$[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

∣ ∣ ∣ \begin{matrix} 5 & - 2 3 & 0 \end{matrix} ∣ ∣ ∣

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

Step 5.2

Find the determinant.

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Step 5.2.1

The determinant of a

2 \times 2

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

∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

5 \cdot 0 - 3 \cdot - 2

$5 \cdot 0 - 3 \cdot - 2$

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

5

$5$ by

0

$0$ .

0 - 3 \cdot - 2

$0 - 3 \cdot - 2$

Step 5.2.2.1.2

Multiply

- 3

$- 3$ by

- 2

$- 2$ .

0 + 6

$0 + 6$

0 + 6

$0 + 6$

Step 5.2.2.2

Add

0

$0$ and

6

$6$ .

6

$6$

6

$6$

D_{y} = 6

$D_{y} = 6$

Step 5.3

Use the formula to solve for

y

$y$ .

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

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

Step 5.4

Substitute

- 17

$- 17$ for

D

$D$ and

6

$6$ for

D_{y}

$D_{y}$ in the formula.

y = \frac{6}{- 17}

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

Step 5.5

Move the negative in front of the fraction.

y = - \frac{6}{17}

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

y = - \frac{6}{17}

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

Step 6

List the solution to the system of equations.

x = - \frac{8}{17}

$x = - \frac{8}{17}$

y = - \frac{6}{17}

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

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