Solve Using a Matrix with Cramer's Rule 5x+3y=3 2x-6y=30

$5 x + 3 y = 3$ $2 x - 6 y = 30$

Represent the system of equations in matrix format.

$[\begin{array}{c} 5 & 3 \\ 2 & - 6 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 3 \\ 30 \end{array}]$

Find the determinant of $[\begin{array}{c} 5 & 3 \\ 2 & - 6 \end{array}]$ .

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

$determinant [\begin{array}{c} 5 & 3 \\ 2 & - 6 \end{array}] = | \begin{array}{c} 5 & 3 \\ 2 & - 6 \end{array} |$

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

$D = (5) (- 6) - 2 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $- 6$ .

$D = - 30 - 2 \cdot 3$

Multiply $- 2$ by $3$ .

$D = - 30 - 6$

Subtract $6$ from $- 30$ .

$D = - 36$

Find the determinant of $[\begin{array}{c} 3 & 3 \\ 30 & - 6 \end{array}]$ .

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

$determinant [\begin{array}{c} 3 & 3 \\ 30 & - 6 \end{array}] = | \begin{array}{c} 3 & 3 \\ 30 & - 6 \end{array} |$

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

$D_{x} = (3) (- 6) - 30 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $- 6$ .

$D_{x} = - 18 - 30 \cdot 3$

Multiply $- 30$ by $3$ .

$D_{x} = - 18 - 90$

Subtract $90$ from $- 18$ .

$D_{x} = - 108$

Find the determinant of $[\begin{array}{c} 5 & 3 \\ 2 & 30 \end{array}]$ .

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

$determinant [\begin{array}{c} 5 & 3 \\ 2 & 30 \end{array}] = | \begin{array}{c} 5 & 3 \\ 2 & 30 \end{array} |$

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

$D_{y} = (5) (30) - 2 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $30$ .

$D_{y} = 150 - 2 \cdot 3$

Multiply $- 2$ by $3$ .

$D_{y} = 150 - 6$

Subtract $6$ from $150$ .

$D_{y} = 144$

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 108}{- 36}$ .

Remove parentheses.

$x = \frac{- 108}{- 36}$

Divide $- 108$ by $- 36$ .

$x = 3$

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{144}{- 36}$ .

Remove parentheses.

$y = \frac{144}{- 36}$

Divide $144$ by $- 36$ .

$y = - 4$

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

$x = 3, y = - 4$

Solve Using a Matrix with Cramer's Rule 5x+3y=3 2x-6y=30

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