Finite Math Examples

$x + 3 y - 13 \cdot - 2 = 18$ , $y - 5 \cdot - 2 = 3$

Step 1

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

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Step 1.1

Multiply $- 13$ by $- 2$ .

$x + 3 y + 26 = 18$

$y - 5 \cdot - 2 = 3$

Step 1.2

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

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Step 1.2.1

Subtract $26$ from both sides of the equation.

$x + 3 y = 18 - 26$

$y - 5 \cdot - 2 = 3$

Step 1.2.2

Subtract $26$ from $18$ .

$x + 3 y = - 8$

$y - 5 \cdot - 2 = 3$

$x + 3 y = - 8$

$y - 5 \cdot - 2 = 3$

Step 1.3

Multiply $- 5$ by $- 2$ .

$x + 3 y = - 8$

$y + 10 = 3$

Step 1.4

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

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Step 1.4.1

Subtract $10$ from both sides of the equation.

$x + 3 y = - 8$

$y = 3 - 10$

Step 1.4.2

Subtract $10$ from $3$ .

$x + 3 y = - 8$

$y = - 7$

$x + 3 y = - 8$

$y = - 7$

$x + 3 y = - 8$

$y = - 7$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 3 \\ 0 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 8 \\ - 7 \end{array}]$

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} 1 & 3 \\ 0 & 1 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} 1 & 3 \\ 0 & 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} 1 & 3 \\ 0 & 1 \end{array} |$

Step 3.2

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

$1 \cdot 1 + 0 \cdot 3$

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 $1$ by $1$ .

$1 + 0 \cdot 3$

Step 3.3.1.2

Multiply $0$ by $3$ .

$1 + 0$

Step 3.3.2

Add $1$ and $0$ .

$1$

$D = 1$

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 $[\begin{array}{c} - 8 \\ - 7 \end{array}]$ .

$| \begin{array}{c} - 8 & 3 \\ - 7 & 1 \end{array} |$

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

$- 8 \cdot 1 - (- 7 \cdot 3)$

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 $- 8$ by $1$ .

$- 8 - (- 7 \cdot 3)$

Step 5.2.2.1.2

Multiply $- (- 7 \cdot 3)$ .

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Step 5.2.2.1.2.1

Multiply $- 7$ by $3$ .

$- 8 - - 21$

Step 5.2.2.1.2.2

Multiply $- 1$ by $- 21$ .

$- 8 + 21$

Step 5.2.2.2

Add $- 8$ and $21$ .

$13$

$D_{x} = 13$

Step 5.3

Use the formula to solve for $x$ .

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

Step 5.4

Substitute $1$ for $D$ and $13$ for $D_{x}$ in the formula.

$x = \frac{13}{1}$

Step 5.5

Divide $13$ by $1$ .

$x = 13$

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 $[\begin{array}{c} - 8 \\ - 7 \end{array}]$ .

$| \begin{array}{c} 1 & - 8 \\ 0 & - 7 \end{array} |$

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

$1 \cdot - 7 + 0 \cdot - 8$

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 $- 7$ by $1$ .

$- 7 + 0 \cdot - 8$

Step 6.2.2.1.2

Multiply $0$ by $- 8$ .

$- 7 + 0$

Step 6.2.2.2

Add $- 7$ and $0$ .

$- 7$

$D_{y} = - 7$

Step 6.3

Use the formula to solve for $y$ .

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

Step 6.4

Substitute $1$ for $D$ and $- 7$ for $D_{y}$ in the formula.

$y = \frac{- 7}{1}$

Step 6.5

Divide $- 7$ by $1$ .

$y = - 7$

Step 7

List the solution to the system of equations.

$x = 13$

$y = - 7$