Solve Using a Matrix with Cramer's Rule -2x+3y=2 , 5x-3y=8

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

Represent the system of equations in matrix format.

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

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

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

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

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

$D = (- 2) (- 3) - 5 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $- 3$ .

$D = 6 - 5 \cdot 3$

Multiply $- 5$ by $3$ .

$D = 6 - 15$

Subtract $15$ from $6$ .

$D = - 9$

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

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

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

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

$D_{x} = (2) (- 3) - 8 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $- 3$ .

$D_{x} = - 6 - 8 \cdot 3$

Multiply $- 8$ by $3$ .

$D_{x} = - 6 - 24$

Subtract $24$ from $- 6$ .

$D_{x} = - 30$

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

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

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $- 2$ by $8$ .

$D_{y} = - 16 - 5 \cdot 2$

Multiply $- 5$ by $2$ .

$D_{y} = - 16 - 10$

Subtract $10$ from $- 16$ .

$D_{y} = - 26$

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 30}{- 9}$ .

Remove parentheses.

$x = \frac{- 30}{- 9}$

Cancel the common factor of $- 30$ and $- 9$ .

Factor $- 3$ out of $- 30$ .

$x = \frac{- 3 (10)}{- 9}$

Cancel the common factors.

Factor $- 3$ out of $- 9$ .

$x = \frac{- 3 \cdot 10}{- 3 \cdot 3}$

Cancel the common factor.

$x = \frac{- 3 \cdot 10}{- 3 \cdot 3}$

Rewrite the expression.

$x = \frac{10}{3}$

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 26}{- 9}$ .

Remove parentheses.

$y = \frac{- 26}{- 9}$

Dividing two negative values results in a positive value.

$y = \frac{26}{9}$

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

$x = \frac{10}{3}, y = \frac{26}{9}$

Solve Using a Matrix with Cramer's Rule -2x+3y=2 , 5x-3y=8

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