Precalculus Examples

$r + \frac{2}{3} s = 4$ , $r - s = 3$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & \frac{2}{3} \\ 1 & - 1 \end{array}] [\begin{array}{c} r \\ s \end{array}] = [\begin{array}{c} 4 \\ 3 \end{array}]$

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Simplify the determinant.

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Step 2.3.1

Multiply $- 1$ by $1$ .

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

Step 2.3.2

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

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

Step 2.3.3

Combine $- 1$ and $\frac{3}{3}$ .

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

Step 2.3.4

Combine the numerators over the common denominator.

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

Step 2.3.5

Simplify the numerator.

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Step 2.3.5.1

Multiply $- 1$ by $3$ .

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

Step 2.3.5.2

Subtract $2$ from $- 3$ .

$\frac{- 5}{3}$

Step 2.3.6

Move the negative in front of the fraction.

$- \frac{5}{3}$

$D = - \frac{5}{3}$

Step 3

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

Step 4

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

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Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $r$ -coefficients of the system with $[\begin{array}{c} 4 \\ 3 \end{array}]$ .

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

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

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

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 $4$ by $- 1$ .

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

Step 4.2.2.1.2

Cancel the common factor of $3$ .

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Step 4.2.2.1.2.1

Factor $3$ out of $- 3$ .

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

Step 4.2.2.1.2.2

Cancel the common factor.

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

Step 4.2.2.1.2.3

Rewrite the expression.

$- 4 - 1 \cdot 2$

Step 4.2.2.1.3

Multiply $- 1$ by $2$ .

$- 4 - 2$

Step 4.2.2.2

Subtract $2$ from $- 4$ .

$- 6$

$D_{r} = - 6$

Step 4.3

Use the formula to solve for $r$ .

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

Step 4.4

Substitute $- \frac{5}{3}$ for $D$ and $- 6$ for $D_{r}$ in the formula.

$r = \frac{- 6}{- \frac{5}{3}}$

Step 4.5

Multiply the numerator by the reciprocal of the denominator.

$r = - 6 (- \frac{3}{5})$

Step 4.6

Multiply $- 6 (- \frac{3}{5})$ .

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Step 4.6.1

Multiply $- 1$ by $- 6$ .

$r = 6 (\frac{3}{5})$

Step 4.6.2

Combine $6$ and $\frac{3}{5}$ .

$r = \frac{6 \cdot 3}{5}$

Step 4.6.3

Multiply $6$ by $3$ .

$r = \frac{18}{5}$

Step 5

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

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Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $s$ -coefficients of the system with $[\begin{array}{c} 4 \\ 3 \end{array}]$ .

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

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

$1 \cdot 3 - 1 \cdot 4$

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 $3$ by $1$ .

$3 - 1 \cdot 4$

Step 5.2.2.1.2

Multiply $- 1$ by $4$ .

$3 - 4$

Step 5.2.2.2

Subtract $4$ from $3$ .

$- 1$

$D_{s} = - 1$

Step 5.3

Use the formula to solve for $s$ .

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

Step 5.4

Substitute $- \frac{5}{3}$ for $D$ and $- 1$ for $D_{s}$ in the formula.

$s = \frac{- 1}{- \frac{5}{3}}$

Step 5.5

Dividing two negative values results in a positive value.

$s = \frac{1}{\frac{5}{3}}$

Step 5.6

Multiply the numerator by the reciprocal of the denominator.

$s = 1 (\frac{3}{5})$

Step 5.7

Multiply $\frac{3}{5}$ by $1$ .

$s = \frac{3}{5}$

Step 6

List the solution to the system of equations.

$r = \frac{18}{5}$

$s = \frac{3}{5}$