Precalculus Examples

$6 x - 5 y = - 1$ , $2 y - 2 x = 0$

Step 1

Reorder $2 y$ and $- 2 x$ .

$6 x - 5 y = - 1$

$- 2 x + 2 y = 0$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 6 & - 5 \\ - 2 & 2 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 1 \\ 0 \end{array}]$

Step 3

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

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Step 3.1

Write $[\begin{array}{c} 6 & - 5 \\ - 2 & 2 \end{array}]$ in determinant notation.

$| \begin{array}{c} 6 & - 5 \\ - 2 & 2 \end{array} |$

Step 3.2

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

$6 \cdot 2 - (- 2 \cdot - 5)$

Step 3.3

Simplify the determinant.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply $6$ by $2$ .

$12 - (- 2 \cdot - 5)$

Step 3.3.1.2

Multiply $- (- 2 \cdot - 5)$ .

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Step 3.3.1.2.1

Multiply $- 2$ by $- 5$ .

$12 - 1 \cdot 10$

Step 3.3.1.2.2

Multiply $- 1$ by $10$ .

$12 - 10$

Step 3.3.2

Subtract $10$ from $12$ .

$2$

$D = 2$

Step 4

Since the determinant is not $0$ , the system can be solved using Cramer's Rule.

Step 5

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ .

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Step 5.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 1 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 1 & - 5 \\ 0 & 2 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

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

$- 1 \cdot 2 + 0 \cdot - 5$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply $- 1$ by $2$ .

$- 2 + 0 \cdot - 5$

Step 5.2.2.1.2

Multiply $0$ by $- 5$ .

$- 2 + 0$

Step 5.2.2.2

Add $- 2$ and $0$ .

$- 2$

$D_{x} = - 2$

Step 5.3

Use the formula to solve for $x$ .

$x = \frac{D_{x}}{D}$

Step 5.4

Substitute $2$ for $D$ and $- 2$ for $D_{x}$ in the formula.

$x = \frac{- 2}{2}$

Step 5.5

Divide $- 2$ by $2$ .

$x = - 1$

Step 6

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ .

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Step 6.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} - 1 \\ 0 \end{array}]$ .

$| \begin{array}{c} 6 & - 1 \\ - 2 & 0 \end{array} |$

Step 6.2

Find the determinant.

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Step 6.2.1

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

$6 \cdot 0 - (- 2 \cdot - 1)$

Step 6.2.2

Simplify the determinant.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Multiply $6$ by $0$ .

$0 - (- 2 \cdot - 1)$

Step 6.2.2.1.2

Multiply $- (- 2 \cdot - 1)$ .

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Step 6.2.2.1.2.1

Multiply $- 2$ by $- 1$ .

$0 - 1 \cdot 2$

Step 6.2.2.1.2.2

Multiply $- 1$ by $2$ .

$0 - 2$

Step 6.2.2.2

Subtract $2$ from $0$ .

$- 2$

$D_{y} = - 2$

Step 6.3

Use the formula to solve for $y$ .

$y = \frac{D_{y}}{D}$

Step 6.4

Substitute $2$ for $D$ and $- 2$ for $D_{y}$ in the formula.

$y = \frac{- 2}{2}$

Step 6.5

Divide $- 2$ by $2$ .

$y = - 1$

Step 7

List the solution to the system of equations.

$x = - 1$

$y = - 1$