Solve Using a Matrix with Cramer's Rule 2x+4y=8 , 6x+3y=-3

$2 x + 4 y = 8$ , $6 x + 3 y = - 3$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 2 & 4 \\ 6 & 3 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 8 \\ - 3 \end{array}]$

Step 2

Find the determinant of $[\begin{array}{c} 2 & 4 \\ 6 & 3 \end{array}]$ .

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

$determinant [\begin{array}{c} 2 & 4 \\ 6 & 3 \end{array}] = | \begin{array}{c} 2 & 4 \\ 6 & 3 \end{array} |$

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

$D = (2) (3) - 6 \cdot 4$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $3$ .

$D = 6 - 6 \cdot 4$

Multiply $- 6$ by $4$ .

$D = 6 - 24$

Subtract $24$ from $6$ .

$D = - 18$

Step 3

Find the determinant of $[\begin{array}{c} 8 & 4 \\ - 3 & 3 \end{array}]$ .

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

$determinant [\begin{array}{c} 8 & 4 \\ - 3 & 3 \end{array}] = | \begin{array}{c} 8 & 4 \\ - 3 & 3 \end{array} |$

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

$D_{x} = (8) (3) + 3 \cdot 4$

Simplify the determinant.

Simplify each term.

Multiply $8$ by $3$ .

$D_{x} = 24 + 3 \cdot 4$

Multiply $3$ by $4$ .

$D_{x} = 24 + 12$

Add $24$ and $12$ .

$D_{x} = 36$

Step 4

Find the determinant of $[\begin{array}{c} 2 & 8 \\ 6 & - 3 \end{array}]$ .

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

$determinant [\begin{array}{c} 2 & 8 \\ 6 & - 3 \end{array}] = | \begin{array}{c} 2 & 8 \\ 6 & - 3 \end{array} |$

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

$D_{y} = (2) (- 3) - 6 \cdot 8$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $- 3$ .

$D_{y} = - 6 - 6 \cdot 8$

Multiply $- 6$ by $8$ .

$D_{y} = - 6 - 48$

Subtract $48$ from $- 6$ .

$D_{y} = - 54$

Step 5

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{36}{- 18}$ .

Remove parentheses.

$x = \frac{36}{- 18}$

Divide $36$ by $- 18$ .

$x = - 2$

Step 6

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 54}{- 18}$ .

Remove parentheses.

$y = \frac{- 54}{- 18}$

Divide $- 54$ by $- 18$ .

$y = 3$

Step 7

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

$x = - 2, y = 3$

Step 8

Solve Using a Matrix with Cramer's Rule 2x+4y=8 , 6x+3y=-3

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