Solve Using a Matrix with Cramer's Rule 3x+2y=2 , 6x+5y=1

$3 x + 2 y = 2$ , $6 x + 5 y = 1$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 3 & 2 \\ 6 & 5 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 2 \\ 1 \end{array}]$

Step 2

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

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

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

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

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

Simplify the determinant.

Simplify each term.

Multiply $3$ by $5$ .

$D = 15 - 6 \cdot 2$

Multiply $- 6$ by $2$ .

$D = 15 - 12$

Subtract $12$ from $15$ .

$D = 3$

Step 3

Find the determinant of $[\begin{array}{c} 2 & 2 \\ 1 & 5 \end{array}]$ .

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

$determinant [\begin{array}{c} 2 & 2 \\ 1 & 5 \end{array}] = | \begin{array}{c} 2 & 2 \\ 1 & 5 \end{array} |$

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

$D_{x} = (2) (5) - 1 \cdot 2$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $5$ .

$D_{x} = 10 - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$D_{x} = 10 - 2$

Subtract $2$ from $10$ .

$D_{x} = 8$

Step 4

Find the determinant of $[\begin{array}{c} 3 & 2 \\ 6 & 1 \end{array}]$ .

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

$determinant [\begin{array}{c} 3 & 2 \\ 6 & 1 \end{array}] = | \begin{array}{c} 3 & 2 \\ 6 & 1 \end{array} |$

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

$D_{y} = (3) (1) - 6 \cdot 2$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $1$ .

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

Multiply $- 6$ by $2$ .

$D_{y} = 3 - 12$

Subtract $12$ from $3$ .

$D_{y} = - 9$

Step 5

Remove parentheses.

$x = \frac{8}{3}$

Step 6

Find the value of $y$ by Cramer\'s Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{- 9}{3}$ .

Remove parentheses.

$y = \frac{- 9}{3}$

Divide $- 9$ by $3$ .

$y = - 3$

Step 7

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

$x = \frac{8}{3}, y = - 3$

Step 8

Solve Using a Matrix with Cramer's Rule 3x+2y=2 , 6x+5y=1

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