Finite Math Examples

y = 4 x + 3 x - 2

$y = 4 x + 3 x - 2$ ,

y = 5 x

$y = 5 x$

Step 1

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

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Step 1.1

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

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Step 1.1.1

Subtract

4 x

$4 x$ from both sides of the equation.

y - 4 x = 3 x - 2

$y - 4 x = 3 x - 2$

y = 5 x

$y = 5 x$

Step 1.1.2

Subtract

3 x

$3 x$ from both sides of the equation.

y - 4 x - 3 x = - 2

$y - 4 x - 3 x = - 2$

y = 5 x

$y = 5 x$

y - 4 x - 3 x = - 2

$y - 4 x - 3 x = - 2$

y = 5 x

$y = 5 x$

Step 1.2

Subtract

3 x

$3 x$ from

- 4 x

$- 4 x$ .

y - 7 x = - 2

$y - 7 x = - 2$

y = 5 x

$y = 5 x$

Step 1.3

Reorder

y

$y$ and

- 7 x

$- 7 x$ .

- 7 x + y = - 2

$- 7 x + y = - 2$

y = 5 x

$y = 5 x$

Step 1.4

Subtract

5 x

$5 x$ from both sides of the equation.

- 7 x + y = - 2

$- 7 x + y = - 2$

y - 5 x = 0

$y - 5 x = 0$

Step 1.5

Reorder

y

$y$ and

- 5 x

$- 5 x$ .

- 7 x + y = - 2

$- 7 x + y = - 2$

- 5 x + y = 0

$- 5 x + y = 0$

- 7 x + y = - 2

$- 7 x + y = - 2$

- 5 x + y = 0

$- 5 x + y = 0$

Step 2

Represent the system of equations in matrix format.

[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]

$[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]$

Step 3

Find the determinant of the coefficient matrix

[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}]

$[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}]$ .

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Step 3.1

Write

[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}]

$[\begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array}]$ in determinant notation.

| \begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array} |

$| \begin{array}{c} - 7 & 1 \\ - 5 & 1 \end{array} |$

Step 3.2

The determinant of a

2 \times 2

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

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

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

- 7 \cdot 1 - (- 5 \cdot 1)

$- 7 \cdot 1 - (- 5 \cdot 1)$

Step 3.3

Simplify the determinant.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply

- 7

$- 7$ by

1

$1$ .

- 7 - (- 5 \cdot 1)

$- 7 - (- 5 \cdot 1)$

Step 3.3.1.2

Multiply

- (- 5 \cdot 1)

$- (- 5 \cdot 1)$ .

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Step 3.3.1.2.1

Multiply

- 5

$- 5$ by

1

$1$ .

- 7 - - 5

$- 7 - - 5$

Step 3.3.1.2.2

Multiply

- 1

$- 1$ by

- 5

$- 5$ .

- 7 + 5

$- 7 + 5$

- 7 + 5

$- 7 + 5$

- 7 + 5

$- 7 + 5$

Step 3.3.2

Add

- 7

$- 7$ and

5

$5$ .

- 2

$- 2$

- 2

$- 2$

D = - 2

$D = - 2$

Step 4

Since the determinant is not

0

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

Step 5

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 5.1

Replace column

1

$1$ of the coefficient matrix that corresponds to the

x

$x$ -coefficients of the system with

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

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

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

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

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

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

- 2 \cdot 1 + 0 \cdot 1

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

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

- 2

$- 2$ by

1

$1$ .

- 2 + 0 \cdot 1

$- 2 + 0 \cdot 1$

Step 5.2.2.1.2

Multiply

0

$0$ by

1

$1$ .

- 2 + 0

$- 2 + 0$

- 2 + 0

$- 2 + 0$

Step 5.2.2.2

Add

- 2

$- 2$ and

0

$0$ .

- 2

$- 2$

- 2

$- 2$

D_{x} = - 2

$D_{x} = - 2$

Step 5.3

Use the formula to solve for

x

$x$ .

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

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

Step 5.4

Substitute

- 2

$- 2$ for

D

$D$ and

- 2

$- 2$ for

D_{x}

$D_{x}$ in the formula.

x = \frac{- 2}{- 2}

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

Step 5.5

Divide

- 2

$- 2$ by

- 2

$- 2$ .

x = 1

$x = 1$

x = 1

$x = 1$

Step 6

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 6.1

Replace column

2

$2$ of the coefficient matrix that corresponds to the

y

$y$ -coefficients of the system with

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

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

| \begin{array}{c} - 7 & - 2 \\ - 5 & 0 \end{array} |

$| \begin{array}{c} - 7 & - 2 \\ - 5 & 0 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

The determinant of a

2 \times 2

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

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

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

- 7 \cdot 0 - (- 5 \cdot - 2)

$- 7 \cdot 0 - (- 5 \cdot - 2)$

Step 6.2.2

Simplify the determinant.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Multiply

- 7

$- 7$ by

0

$0$ .

0 - (- 5 \cdot - 2)

$0 - (- 5 \cdot - 2)$

Step 6.2.2.1.2

Multiply

- (- 5 \cdot - 2)

$- (- 5 \cdot - 2)$ .

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Step 6.2.2.1.2.1

Multiply

- 5

$- 5$ by

- 2

$- 2$ .

0 - 1 \cdot 10

$0 - 1 \cdot 10$

Step 6.2.2.1.2.2

Multiply

- 1

$- 1$ by

10

$10$ .

0 - 10

$0 - 10$

0 - 10

$0 - 10$

0 - 10

$0 - 10$

Step 6.2.2.2

Subtract

10

$10$ from

0

$0$ .

- 10

$- 10$

- 10

$- 10$

D_{y} = - 10

$D_{y} = - 10$

Step 6.3

Use the formula to solve for

y

$y$ .

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

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

Step 6.4

Substitute

- 2

$- 2$ for

D

$D$ and

- 10

$- 10$ for

D_{y}

$D_{y}$ in the formula.

y = \frac{- 10}{- 2}

$y = \frac{- 10}{- 2}$

Step 6.5

Divide

- 10

$- 10$ by

- 2

$- 2$ .

y = 5

$y = 5$

y = 5

$y = 5$

Step 7

List the solution to the system of equations.

x = 1

$x = 1$

y = 5

$y = 5$