Finite Math Examples

$x - 6 y = 3$ , $x - y = - 1$

Step 1

Represent the system of equations in matrix format.

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

$1 \cdot - 1 - 1 \cdot - 6$

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 $- 1$ by $1$ .

$- 1 - 1 \cdot - 6$

Step 2.3.1.2

Multiply $- 1$ by $- 6$ .

$- 1 + 6$

Step 2.3.2

Add $- 1$ and $6$ .

$5$

$D = 5$

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 $[\begin{array}{c} 3 \\ - 1 \end{array}]$ .

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

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

$3 \cdot - 1 - - - 6$

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 $3$ by $- 1$ .

$- 3 - - - 6$

Step 4.2.2.1.2

Multiply $- - - 6$ .

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Step 4.2.2.1.2.1

Multiply $- 1$ by $- 6$ .

$- 3 - 1 \cdot 6$

Step 4.2.2.1.2.2

Multiply $- 1$ by $6$ .

$- 3 - 6$

Step 4.2.2.2

Subtract $6$ from $- 3$ .

$- 9$

$D_{x} = - 9$

Step 4.3

Use the formula to solve for $x$ .

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

Step 4.4

Substitute $5$ for $D$ and $- 9$ for $D_{x}$ in the formula.

$x = \frac{- 9}{5}$

Step 4.5

Move the negative in front of the fraction.

$x = - \frac{9}{5}$

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 $[\begin{array}{c} 3 \\ - 1 \end{array}]$ .

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

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

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

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 $1$ .

$- 1 - 1 \cdot 3$

Step 5.2.2.1.2

Multiply $- 1$ by $3$ .

$- 1 - 3$

Step 5.2.2.2

Subtract $3$ from $- 1$ .

$- 4$

$D_{y} = - 4$

Step 5.3

Use the formula to solve for $y$ .

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

Step 5.4

Substitute $5$ for $D$ and $- 4$ for $D_{y}$ in the formula.

$y = \frac{- 4}{5}$

Step 5.5

Move the negative in front of the fraction.

$y = - \frac{4}{5}$

Step 6

List the solution to the system of equations.

$x = - \frac{9}{5}$

$y = - \frac{4}{5}$

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