Precalculus Examples

$- (y - 4) = x + 9$ , $x - \frac{8}{3} y = 0$

Step 1

Move all of the variables to the left side of each equation.

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Step 1.1

Subtract $x$ from both sides of the equation.

$- (y - 4) - x = 9$

$x - \frac{8}{3} y = 0$

Step 1.2

Simplify each term.

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Step 1.2.1

Apply the distributive property.

$- y - - 4 - x = 9$

$x - \frac{8}{3} y = 0$

Step 1.2.2

Multiply $- 1$ by $- 4$ .

$- y + 4 - x = 9$

$x - \frac{8}{3} y = 0$

$- y + 4 - x = 9$

$x - \frac{8}{3} y = 0$

Step 1.3

Move all terms not containing a variable to the right side of the equation.

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Step 1.3.1

Subtract $4$ from both sides of the equation.

$- y - x = 9 - 4$

$x - \frac{8}{3} y = 0$

Step 1.3.2

Subtract $4$ from $9$ .

$- y - x = 5$

$x - \frac{8}{3} y = 0$

$- y - x = 5$

$x - \frac{8}{3} y = 0$

Step 1.4

Reorder $- y$ and $- x$ .

$- x - y = 5$

$x - \frac{8}{3} y = 0$

Step 1.5

Simplify each term.

$- x - y = 5$

$x - \frac{8 y}{3} = 0$

Step 1.6

Reorder terms.

$- x - y = 5$

$x - (\frac{8}{3} y) = 0$

Step 1.7

Remove parentheses.

$- x - y = 5$

$x - \frac{8}{3} y = 0$

$- x - y = 5$

$x - \frac{8}{3} y = 0$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} - 1 & - 1 \\ 1 & - \frac{8}{3} \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 5 \\ 0 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} - 1 & - 1 \\ 1 & - \frac{8}{3} \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} - 1 & - 1 \\ 1 & - \frac{8}{3} \end{array}]$ in determinant notation.

$| \begin{array}{c} - 1 & - 1 \\ 1 & - \frac{8}{3} \end{array} |$

Step 3.2

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

$- - \frac{8}{3} - 1 \cdot - 1$

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 $- - \frac{8}{3}$ .

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Step 3.3.1.1.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{8}{3}) - 1 \cdot - 1$

Step 3.3.1.1.2

Multiply $\frac{8}{3}$ by $1$ .

$\frac{8}{3} - 1 \cdot - 1$

Step 3.3.1.2

Multiply $- 1$ by $- 1$ .

$\frac{8}{3} + 1$

Step 3.3.2

Write $1$ as a fraction with a common denominator.

$\frac{8}{3} + \frac{3}{3}$

Step 3.3.3

Combine the numerators over the common denominator.

$\frac{8 + 3}{3}$

Step 3.3.4

Add $8$ and $3$ .

$\frac{11}{3}$

$D = \frac{11}{3}$

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 $[\begin{array}{c} 5 \\ 0 \end{array}]$ .

$| \begin{array}{c} 5 & - 1 \\ 0 & - \frac{8}{3} \end{array} |$

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

$5 (- \frac{8}{3}) + 0 \cdot - 1$

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 $5 (- \frac{8}{3})$ .

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Step 5.2.2.1.1.1

Multiply $- 1$ by $5$ .

$- 5 (\frac{8}{3}) + 0 \cdot - 1$

Step 5.2.2.1.1.2

Combine $- 5$ and $\frac{8}{3}$ .

$\frac{- 5 \cdot 8}{3} + 0 \cdot - 1$

Step 5.2.2.1.1.3

Multiply $- 5$ by $8$ .

$\frac{- 40}{3} + 0 \cdot - 1$

Step 5.2.2.1.2

Move the negative in front of the fraction.

$- \frac{40}{3} + 0 \cdot - 1$

Step 5.2.2.1.3

Multiply $0$ by $- 1$ .

$- \frac{40}{3} + 0$

Step 5.2.2.2

Add $- \frac{40}{3}$ and $0$ .

$- \frac{40}{3}$

$D_{x} = - \frac{40}{3}$

Step 5.3

Use the formula to solve for $x$ .

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

Step 5.4

Substitute $\frac{11}{3}$ for $D$ and $- \frac{40}{3}$ for $D_{x}$ in the formula.

$x = \frac{- \frac{40}{3}}{\frac{11}{3}}$

Step 5.5

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{40}{3} \cdot \frac{3}{11}$

Step 5.6

Cancel the common factor of $3$ .

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Step 5.6.1

Move the leading negative in $- \frac{40}{3}$ into the numerator.

$x = \frac{- 40}{3} \cdot \frac{3}{11}$

Step 5.6.2

Cancel the common factor.

$x = \frac{- 40}{3} \cdot \frac{3}{11}$

Step 5.6.3

Rewrite the expression.

$x = - 40 (\frac{1}{11})$

Step 5.7

Combine $- 40$ and $\frac{1}{11}$ .

$x = \frac{- 40}{11}$

Step 5.8

Move the negative in front of the fraction.

$x = - \frac{40}{11}$

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 $[\begin{array}{c} 5 \\ 0 \end{array}]$ .

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

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

$- 0 - 1 \cdot 5$

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 $- 1$ by $0$ .

$0 - 1 \cdot 5$

Step 6.2.2.1.2

Multiply $- 1$ by $5$ .

$0 - 5$

Step 6.2.2.2

Subtract $5$ from $0$ .

$- 5$

$D_{y} = - 5$

Step 6.3

Use the formula to solve for $y$ .

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

Step 6.4

Substitute $\frac{11}{3}$ for $D$ and $- 5$ for $D_{y}$ in the formula.

$y = \frac{- 5}{\frac{11}{3}}$

Step 6.5

Multiply the numerator by the reciprocal of the denominator.

$y = - 5 (\frac{3}{11})$

Step 6.6

Multiply $- 5 (\frac{3}{11})$ .

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Step 6.6.1

Combine $- 5$ and $\frac{3}{11}$ .

$y = \frac{- 5 \cdot 3}{11}$

Step 6.6.2

Multiply $- 5$ by $3$ .

$y = \frac{- 15}{11}$

Step 6.7

Move the negative in front of the fraction.

$y = - \frac{15}{11}$

Step 7

List the solution to the system of equations.

$x = - \frac{40}{11}$

$y = - \frac{15}{11}$