Finite Math Examples

$- x + 7 y = 35$ , $3 x - 4 y = - 5$

Represent the system of equations in matrix format.

$[\begin{array}{c} - 1 & 7 \\ 3 & - 4 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 35 \\ - 5 \end{array}]$

The determinant of $[\begin{array}{c} - 1 & 7 \\ 3 & - 4 \end{array}]$ is $- 17$ .

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

$determinant [\begin{array}{c} - 1 & 7 \\ 3 & - 4 \end{array}] = | \begin{array}{c} - 1 & 7 \\ 3 & - 4 \end{array} |$

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

$D = (- 1) (- 4) - 3 \cdot 7$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $- 4$ .

$D = 4 - 3 \cdot 7$

Multiply $- 3$ by $7$ .

$D = 4 - 21$

Subtract $21$ from $4$ .

$D = - 17$

The determinant of $[\begin{array}{c} 35 & 7 \\ - 5 & - 4 \end{array}]$ is $- 105$ .

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

$determinant [\begin{array}{c} 35 & 7 \\ - 5 & - 4 \end{array}] = | \begin{array}{c} 35 & 7 \\ - 5 & - 4 \end{array} |$

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

$D_{x} = (35) (- 4) + 5 \cdot 7$

Simplify the determinant.

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Simplify each term.

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Multiply $35$ by $- 4$ .

$D_{x} = - 140 + 5 \cdot 7$

Multiply $5$ by $7$ .

$D_{x} = - 140 + 35$

Add $- 140$ and $35$ .

$D_{x} = - 105$

The determinant of $[\begin{array}{c} - 1 & 35 \\ 3 & - 5 \end{array}]$ is $- 100$ .

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

$determinant [\begin{array}{c} - 1 & 35 \\ 3 & - 5 \end{array}] = | \begin{array}{c} - 1 & 35 \\ 3 & - 5 \end{array} |$

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

$D_{y} = (- 1) (- 5) - 3 \cdot 35$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $- 5$ .

$D_{y} = 5 - 3 \cdot 35$

Multiply $- 3$ by $35$ .

$D_{y} = 5 - 105$

Subtract $105$ from $5$ .

$D_{y} = - 100$

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 105}{- 17}$ .

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Remove the parentheses from the numerator.

$x = - \frac{105}{- 17}$

Simplify $- \frac{105}{- 17}$ .

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Move the negative in front of the fraction.

$x = \frac{105}{17}$

Multiply $- - \frac{105}{17}$ .

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Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{105}{17})$

Multiply $\frac{105}{17}$ by $1$ .

$x = \frac{105}{17}$

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 100}{- 17}$ .

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Remove the parentheses from the numerator.

$y = - \frac{100}{- 17}$

Simplify $- \frac{100}{- 17}$ .

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Move the negative in front of the fraction.

$y = \frac{100}{17}$

Multiply $- - \frac{100}{17}$ .

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Multiply $- 1$ by $- 1$ .

$y = 1 (\frac{100}{17})$

Multiply $\frac{100}{17}$ by $1$ .

$y = \frac{100}{17}$

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

$x = \frac{105}{17}, y = \frac{100}{17}$

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