Precalculus Examples

$9 x - y = 0$ , $3 x - y - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

$9 x - y = 0$

$3 x - y = 4$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 9 & - 1 \\ 3 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 0 \\ 4 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} 9 & - 1 \\ 3 & - 1 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} 9 & - 1 \\ 3 & - 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} 9 & - 1 \\ 3 & - 1 \end{array} |$

Step 3.2

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

$9 \cdot - 1 - 3 \cdot - 1$

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 $9$ by $- 1$ .

$- 9 - 3 \cdot - 1$

Step 3.3.1.2

Multiply $- 3$ by $- 1$ .

$- 9 + 3$

Step 3.3.2

Add $- 9$ and $3$ .

$- 6$

$D = - 6$

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 $[\begin{array}{c} 0 \\ 4 \end{array}]$ .

$| \begin{array}{c} 0 & - 1 \\ 4 & - 1 \end{array} |$

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 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$0 \cdot - 1 - 4 \cdot - 1$

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 $0$ by $- 1$ .

$0 - 4 \cdot - 1$

Step 5.2.2.1.2

Multiply $- 4$ by $- 1$ .

$0 + 4$

Step 5.2.2.2

Add $0$ and $4$ .

$4$

$D_{x} = 4$

Step 5.3

Use the formula to solve for $x$ .

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

Step 5.4

Substitute $- 6$ for $D$ and $4$ for $D_{x}$ in the formula.

$x = \frac{4}{- 6}$

Step 5.5

Cancel the common factor of $4$ and $- 6$ .

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Step 5.5.1

Factor $2$ out of $4$ .

$x = \frac{2 (2)}{- 6}$

Step 5.5.2

Cancel the common factors.

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Step 5.5.2.1

Factor $2$ out of $- 6$ .

$x = \frac{2 \cdot 2}{2 \cdot - 3}$

Step 5.5.2.2

Cancel the common factor.

$x = \frac{2 \cdot 2}{2 \cdot - 3}$

Step 5.5.2.3

Rewrite the expression.

$x = \frac{2}{- 3}$

Step 5.6

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

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 $[\begin{array}{c} 0 \\ 4 \end{array}]$ .

$| \begin{array}{c} 9 & 0 \\ 3 & 4 \end{array} |$

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 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$9 \cdot 4 - 3 \cdot 0$

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 $9$ by $4$ .

$36 - 3 \cdot 0$

Step 6.2.2.1.2

Multiply $- 3$ by $0$ .

$36 + 0$

Step 6.2.2.2

Add $36$ and $0$ .

$36$

$D_{y} = 36$

Step 6.3

Use the formula to solve for $y$ .

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

Step 6.4

Substitute $- 6$ for $D$ and $36$ for $D_{y}$ in the formula.

$y = \frac{36}{- 6}$

Step 6.5

Divide $36$ by $- 6$ .

$y = - 6$

Step 7

List the solution to the system of equations.

$x = - \frac{2}{3}$

$y = - 6$