Finite Math Examples

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

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

$5 x + 3 - 4 y = 0$

$y = 8 x - 2$

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

$5 x + 3 - 4 y = 0$

$y - 8 x = - 2$

Subtract $3$ from both sides of the equation.

$5 x - 4 y = - 3$

$y - 8 x = - 2$

Represent the system of equations in matrix format.

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

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

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array}] = | \begin{array}{c} 5 & - 4 \\ - 8 & 1 \end{array} |$

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

$D = (5) (1) + 8 \cdot - 4$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $1$ .

$D = 5 + 8 \cdot - 4$

Multiply $8$ by $- 4$ .

$D = 5 - 32$

Subtract $32$ from $5$ .

$D = - 27$

Find the determinant of $[\begin{array}{c} - 3 & - 4 \\ - 2 & 1 \end{array}]$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} - 3 & - 4 \\ - 2 & 1 \end{array}] = | \begin{array}{c} - 3 & - 4 \\ - 2 & 1 \end{array} |$

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

$D_{x} = (- 3) (1) + 2 \cdot - 4$

Simplify the determinant.

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Simplify each term.

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Multiply $- 3$ by $1$ .

$D_{x} = - 3 + 2 \cdot - 4$

Multiply $2$ by $- 4$ .

$D_{x} = - 3 - 8$

Subtract $8$ from $- 3$ .

$D_{x} = - 11$

Find the determinant of $[\begin{array}{c} 5 & - 3 \\ - 8 & - 2 \end{array}]$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 5 & - 3 \\ - 8 & - 2 \end{array}] = | \begin{array}{c} 5 & - 3 \\ - 8 & - 2 \end{array} |$

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

$D_{y} = (5) (- 2) + 8 \cdot - 3$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $- 2$ .

$D_{y} = - 10 + 8 \cdot - 3$

Multiply $8$ by $- 3$ .

$D_{y} = - 10 - 24$

Subtract $24$ from $- 10$ .

$D_{y} = - 34$

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 11}{- 27}$ .

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Remove parentheses.

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

Dividing two negative values results in a positive value.

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

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 34}{- 27}$ .

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Remove parentheses.

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

Dividing two negative values results in a positive value.

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

The solution to the system of equations using Cramer's Rule.

$x = \frac{11}{27}, y = \frac{34}{27}$

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