Precalculus Examples

$\frac{1}{2} x + \frac{1}{3} y = 1$ , $\frac{1}{4} x - \frac{1}{6} y = - \frac{3}{2}$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} \frac{1}{2} & \frac{1}{3} \\ \frac{1}{4} & - \frac{1}{6} \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ - \frac{3}{2} \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} \frac{1}{2} & \frac{1}{3} \\ \frac{1}{4} & - \frac{1}{6} \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} \frac{1}{2} & \frac{1}{3} \\ \frac{1}{4} & - \frac{1}{6} \end{array}]$ in determinant notation.

$| \begin{array}{c} \frac{1}{2} & \frac{1}{3} \\ \frac{1}{4} & - \frac{1}{6} \end{array} |$

Step 2.2

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

$\frac{1}{2} (- \frac{1}{6}) - \frac{1}{4} \cdot \frac{1}{3}$

Step 2.3

Simplify the determinant.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Multiply $\frac{1}{2} (- \frac{1}{6})$ .

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Step 2.3.1.1.1

Multiply $\frac{1}{2}$ by $\frac{1}{6}$ .

$- \frac{1}{2 \cdot 6} - \frac{1}{4} \cdot \frac{1}{3}$

Step 2.3.1.1.2

Multiply $2$ by $6$ .

$- \frac{1}{12} - \frac{1}{4} \cdot \frac{1}{3}$

Step 2.3.1.2

Multiply $- \frac{1}{4} \cdot \frac{1}{3}$ .

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Step 2.3.1.2.1

Multiply $\frac{1}{3}$ by $\frac{1}{4}$ .

$- \frac{1}{12} - \frac{1}{3 \cdot 4}$

Step 2.3.1.2.2

Multiply $3$ by $4$ .

$- \frac{1}{12} - \frac{1}{12}$

Step 2.3.2

Combine the numerators over the common denominator.

$\frac{- 1 - 1}{12}$

Step 2.3.3

Subtract $1$ from $- 1$ .

$\frac{- 2}{12}$

Step 2.3.4

Cancel the common factor of $- 2$ and $12$ .

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Step 2.3.4.1

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{12}$

Step 2.3.4.2

Cancel the common factors.

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Step 2.3.4.2.1

Factor $2$ out of $12$ .

$\frac{2 \cdot - 1}{2 \cdot 6}$

Step 2.3.4.2.2

Cancel the common factor.

$\frac{2 \cdot - 1}{2 \cdot 6}$

Step 2.3.4.2.3

Rewrite the expression.

$\frac{- 1}{6}$

Step 2.3.5

Move the negative in front of the fraction.

$- \frac{1}{6}$

$D = - \frac{1}{6}$

Step 3

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

Step 4

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

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Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} 1 \\ - \frac{3}{2} \end{array}]$ .

$| \begin{array}{c} 1 & \frac{1}{3} \\ - \frac{3}{2} & - \frac{1}{6} \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

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

$1 (- \frac{1}{6}) - (- \frac{3}{2} \cdot \frac{1}{3})$

Step 4.2.2

Simplify the determinant.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply $- \frac{1}{6}$ by $1$ .

$- \frac{1}{6} - (- \frac{3}{2} \cdot \frac{1}{3})$

Step 4.2.2.1.2

Cancel the common factor of $3$ .

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Step 4.2.2.1.2.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$- \frac{1}{6} - (\frac{- 3}{2} \cdot \frac{1}{3})$

Step 4.2.2.1.2.2

Factor $3$ out of $- 3$ .

$- \frac{1}{6} - (\frac{3 (- 1)}{2} \cdot \frac{1}{3})$

Step 4.2.2.1.2.3

Cancel the common factor.

$- \frac{1}{6} - (\frac{3 \cdot - 1}{2} \cdot \frac{1}{3})$

Step 4.2.2.1.2.4

Rewrite the expression.

$- \frac{1}{6} - \frac{- 1}{2}$

Step 4.2.2.1.3

Move the negative in front of the fraction.

$- \frac{1}{6} - - \frac{1}{2}$

Step 4.2.2.1.4

Multiply $- - \frac{1}{2}$ .

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Step 4.2.2.1.4.1

Multiply $- 1$ by $- 1$ .

$- \frac{1}{6} + 1 (\frac{1}{2})$

Step 4.2.2.1.4.2

Multiply $\frac{1}{2}$ by $1$ .

$- \frac{1}{6} + \frac{1}{2}$

Step 4.2.2.2

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{1}{6} + \frac{1}{2} \cdot \frac{3}{3}$

Step 4.2.2.3

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.2.2.3.1

Multiply $\frac{1}{2}$ by $\frac{3}{3}$ .

$- \frac{1}{6} + \frac{3}{2 \cdot 3}$

Step 4.2.2.3.2

Multiply $2$ by $3$ .

$- \frac{1}{6} + \frac{3}{6}$

Step 4.2.2.4

Combine the numerators over the common denominator.

$\frac{- 1 + 3}{6}$

Step 4.2.2.5

Add $- 1$ and $3$ .

$\frac{2}{6}$

Step 4.2.2.6

Cancel the common factor of $2$ and $6$ .

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Step 4.2.2.6.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{6}$

Step 4.2.2.6.2

Cancel the common factors.

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Step 4.2.2.6.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 1}{2 \cdot 3}$

Step 4.2.2.6.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 3}$

Step 4.2.2.6.2.3

Rewrite the expression.

$\frac{1}{3}$

$D_{x} = \frac{1}{3}$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $- \frac{1}{6}$ for $D$ and $\frac{1}{3}$ for $D_{x}$ in the formula.

$x = \frac{\frac{1}{3}}{- \frac{1}{6}}$

Step 4.5

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{1}{3} (- 1 \cdot 6)$

Step 4.6

Multiply $- 1$ by $6$ .

$x = \frac{1}{3} \cdot - 6$

Step 4.7

Cancel the common factor of $3$ .

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Step 4.7.1

Factor $3$ out of $- 6$ .

$x = \frac{1}{3} \cdot (3 (- 2))$

Step 4.7.2

Cancel the common factor.

$x = \frac{1}{3} \cdot (3 \cdot - 2)$

Step 4.7.3

Rewrite the expression.

$x = - 2$

Step 5

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

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Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} 1 \\ - \frac{3}{2} \end{array}]$ .

$| \begin{array}{c} \frac{1}{2} & 1 \\ \frac{1}{4} & - \frac{3}{2} \end{array} |$

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

$\frac{1}{2} (- \frac{3}{2}) - \frac{1}{4} \cdot 1$

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 $\frac{1}{2} (- \frac{3}{2})$ .

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Step 5.2.2.1.1.1

Multiply $\frac{1}{2}$ by $\frac{3}{2}$ .

$- \frac{3}{2 \cdot 2} - \frac{1}{4} \cdot 1$

Step 5.2.2.1.1.2

Multiply $2$ by $2$ .

$- \frac{3}{4} - \frac{1}{4} \cdot 1$

Step 5.2.2.1.2

Multiply $- 1$ by $1$ .

$- \frac{3}{4} - \frac{1}{4}$

Step 5.2.2.2

Combine the numerators over the common denominator.

$\frac{- 3 - 1}{4}$

Step 5.2.2.3

Subtract $1$ from $- 3$ .

$\frac{- 4}{4}$

Step 5.2.2.4

Divide $- 4$ by $4$ .

$- 1$

$D_{y} = - 1$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $- \frac{1}{6}$ for $D$ and $- 1$ for $D_{y}$ in the formula.

$y = \frac{- 1}{- \frac{1}{6}}$

Step 5.5

Dividing two negative values results in a positive value.

$y = \frac{1}{\frac{1}{6}}$

Step 5.6

Multiply the numerator by the reciprocal of the denominator.

$y = 1 \cdot 6$

Step 5.7

Multiply $6$ by $1$ .

$y = 6$

Step 6

List the solution to the system of equations.

$x = - 2$

$y = 6$