Finite Math Examples

$5 x + 3 = 4 y$ , $y = 8 x - 2$

Step 1

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

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Step 1.1

Subtract $4 y$ from both sides of the equation.

$5 x + 3 - 4 y = 0$

$y = 8 x - 2$

Step 1.2

Subtract $3$ from both sides of the equation.

$5 x - 4 y = - 3$

$y = 8 x - 2$

Step 1.3

Subtract $8 x$ from both sides of the equation.

$5 x - 4 y = - 3$

$y - 8 x = - 2$

Step 1.4

Reorder $y$ and $- 8 x$ .

$5 x - 4 y = - 3$

$- 8 x + y = - 2$

$5 x - 4 y = - 3$

$- 8 x + y = - 2$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 3 \\ - 2 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array} |$

Step 3.2

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

$5 \cdot 1 - (- 8 \cdot - 4)$

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 $5$ by $1$ .

$5 - (- 8 \cdot - 4)$

Step 3.3.1.2

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

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Step 3.3.1.2.1

Multiply $- 8$ by $- 4$ .

$5 - 1 \cdot 32$

Step 3.3.1.2.2

Multiply $- 1$ by $32$ .

$5 - 32$

Step 3.3.2

Subtract $32$ from $5$ .

$- 27$

$D = - 27$

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

$| \begin{array}{c} - 3 & - 4 \\ - 2 & 1 \end{array} |$

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

$- 3 \cdot 1 - (- 2 \cdot - 4)$

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 $- 3$ by $1$ .

$- 3 - (- 2 \cdot - 4)$

Step 5.2.2.1.2

Multiply $- (- 2 \cdot - 4)$ .

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Step 5.2.2.1.2.1

Multiply $- 2$ by $- 4$ .

$- 3 - 1 \cdot 8$

Step 5.2.2.1.2.2

Multiply $- 1$ by $8$ .

$- 3 - 8$

Step 5.2.2.2

Subtract $8$ from $- 3$ .

$- 11$

$D_{x} = - 11$

Step 5.3

Use the formula to solve for $x$ .

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

Step 5.4

Substitute $- 27$ for $D$ and $- 11$ for $D_{x}$ in the formula.

$x = \frac{- 11}{- 27}$

Step 5.5

Dividing two negative values results in a positive value.

$x = \frac{11}{27}$

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

$| \begin{array}{c} 5 & - 3 \\ - 8 & - 2 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

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

$5 \cdot - 2 - (- 8 \cdot - 3)$

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 $5$ by $- 2$ .

$- 10 - (- 8 \cdot - 3)$

Step 6.2.2.1.2

Multiply $- (- 8 \cdot - 3)$ .

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Step 6.2.2.1.2.1

Multiply $- 8$ by $- 3$ .

$- 10 - 1 \cdot 24$

Step 6.2.2.1.2.2

Multiply $- 1$ by $24$ .

$- 10 - 24$

Step 6.2.2.2

Subtract $24$ from $- 10$ .

$- 34$

$D_{y} = - 34$

Step 6.3

Use the formula to solve for $y$ .

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

Step 6.4

Substitute $- 27$ for $D$ and $- 34$ for $D_{y}$ in the formula.

$y = \frac{- 34}{- 27}$

Step 6.5

Dividing two negative values results in a positive value.

$y = \frac{34}{27}$

Step 7

List the solution to the system of equations.

$x = \frac{11}{27}$

$y = \frac{34}{27}$

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