Finite Math Examples

$y = 3 x + z - 2$ , $z = 3 x + 4$ , $y = 5 z$

Move all terms containing variables to the left side of the equation.

Tap for more steps...

Move $3 x$ to the left side of the equation because it contains a variable.

$y - 3 x = z - 2$

$z = 3 x + 4$

$y = 5 z$

Move $z$ to the left side of the equation because it contains a variable.

$y - 3 x - z = - 2$

$z = 3 x + 4$

$y = 5 z$

$y - 3 x - z = - 2$

$z = 3 x + 4$

$y = 5 z$

Move $3 x$ to the left side of the equation because it contains a variable.

$y - 3 x - z = - 2$

$z - 3 x = 4$

$y = 5 z$

Move $5 z$ to the left side of the equation because it contains a variable.

$y - 3 x - z = - 2$

$z - 3 x = 4$

$y - 5 z = 0$

Represent the system of equations in matrix format.

$[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 2 \\ 4 \\ 0 \end{array}]$

The determinant of $[\begin{array}{c} - 3 & 1 & - 1 \\ - 3 & 0 & 1 \\ 0 & 1 & - 5 \end{array}]$ is $- 9$ .

Tap for more steps...

Set up the determinant by breaking it into smaller components.

$D = - 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

The determinant of $- 3 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$ is $3$ .

Tap for more steps...

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

$D = - 3 ((0) (- 5) - 1 \cdot 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $0$ by $- 5$ .

$D = - 3 (0 - 1 \cdot 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 1$ by $1$ .

$D = - 3 (0 - 1) + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the expression.

Tap for more steps...

Subtract $1$ from $0$ .

$D = - 3 \cdot - 1 + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 3$ by $- 1$ .

$D = 3 + 3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

The determinant of $3 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$ is $- 12$ .

Tap for more steps...

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

$D = 3 + 3 ((1) (- 5) - 1 \cdot - 1) + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $- 5$ by $1$ .

$D = 3 + 3 (- 5 - 1 \cdot - 1) + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 1$ by $- 1$ .

$D = 3 + 3 (- 5 + 1) + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the expression.

Tap for more steps...

Add $- 5$ and $1$ .

$D = 3 + 3 \cdot - 4 + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $3$ by $- 4$ .

$D = 3 - 12 + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D = 3 - 12 + 0$

Subtract $12$ from $3$ .

$D = - 9 + 0$

Add $- 9$ and $0$ .

$D = - 9$

The determinant of $[\begin{array}{c} - 2 & 1 & - 1 \\ 4 & 0 & 1 \\ 0 & 1 & - 5 \end{array}]$ is $18$ .

Tap for more steps...

Set up the determinant by breaking it into smaller components.

$D_{x} = - 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} | - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

The determinant of $- 2 | \begin{array}{c} 0 & 1 \\ 1 & - 5 \end{array} |$ is $2$ .

Tap for more steps...

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

$D_{x} = - 2 ((0) (- 5) - 1 \cdot 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $0$ by $- 5$ .

$D_{x} = - 2 (0 - 1 \cdot 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 1$ by $1$ .

$D_{x} = - 2 (0 - 1) - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the expression.

Tap for more steps...

Subtract $1$ from $0$ .

$D_{x} = - 2 \cdot - 1 - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 2$ by $- 1$ .

$D_{x} = 2 - 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} | + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

The determinant of $- 4 | \begin{array}{c} 1 & - 1 \\ 1 & - 5 \end{array} |$ is $16$ .

Tap for more steps...

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

$D_{x} = 2 - 4 ((1) (- 5) - 1 \cdot - 1) + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $- 5$ by $1$ .

$D_{x} = 2 - 4 (- 5 - 1 \cdot - 1) + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 1$ by $- 1$ .

$D_{x} = 2 - 4 (- 5 + 1) + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Simplify the expression.

Tap for more steps...

Add $- 5$ and $1$ .

$D_{x} = 2 - 4 \cdot - 4 + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Multiply $- 4$ by $- 4$ .

$D_{x} = 2 + 16 + 0 | \begin{array}{c} 1 & - 1 \\ 0 & 1 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{x} = 2 + 16 + 0$

Add $2$ and $16$ .

$D_{x} = 18 + 0$

Add $18$ and $0$ .

$D_{x} = 18$

The determinant of $[\begin{array}{c} - 3 & - 2 & - 1 \\ - 3 & 4 & 1 \\ 0 & 0 & - 5 \end{array}]$ is $90$ .

Tap for more steps...

Set up the determinant by breaking it into smaller components.

$D_{y} = - 3 | \begin{array}{c} 4 & 1 \\ 0 & - 5 \end{array} | + 3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} | + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

The determinant of $- 3 | \begin{array}{c} 4 & 1 \\ 0 & - 5 \end{array} |$ is $60$ .

Tap for more steps...

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

$D_{y} = - 3 ((4) (- 5) + 0 \cdot 1) + 3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} | + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $4$ by $- 5$ .

$D_{y} = - 3 (- 20 + 0 \cdot 1) + 3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} | + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Multiply $0$ by $1$ .

$D_{y} = - 3 (- 20 + 0) + 3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} | + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Simplify the expression.

Tap for more steps...

Add $- 20$ and $0$ .

$D_{y} = - 3 \cdot - 20 + 3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} | + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Multiply $- 3$ by $- 20$ .

$D_{y} = 60 + 3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} | + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

The determinant of $3 | \begin{array}{c} - 2 & - 1 \\ 0 & - 5 \end{array} |$ is $30$ .

Tap for more steps...

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

$D_{y} = 60 + 3 ((- 2) (- 5) + 0 \cdot - 1) + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $- 2$ by $- 5$ .

$D_{y} = 60 + 3 (10 + 0 \cdot - 1) + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Multiply $0$ by $- 1$ .

$D_{y} = 60 + 3 (10 + 0) + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Simplify the expression.

Tap for more steps...

Add $10$ and $0$ .

$D_{y} = 60 + 3 \cdot 10 + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Multiply $3$ by $10$ .

$D_{y} = 60 + 30 + 0 | \begin{array}{c} - 2 & - 1 \\ 4 & 1 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{y} = 60 + 30 + 0$

Add $60$ and $30$ .

$D_{y} = 90 + 0$

Add $90$ and $0$ .

$D_{y} = 90$

The determinant of $[\begin{array}{c} - 3 & 1 & - 2 \\ - 3 & 0 & 4 \\ 0 & 1 & 0 \end{array}]$ is $18$ .

Tap for more steps...

Set up the determinant by breaking it into smaller components.

$D_{z} = - 3 | \begin{array}{c} 0 & 4 \\ 1 & 0 \end{array} | + 3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} | + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

The determinant of $- 3 | \begin{array}{c} 0 & 4 \\ 1 & 0 \end{array} |$ is $12$ .

Tap for more steps...

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

$D_{z} = - 3 ((0) (0) - 1 \cdot 4) + 3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} | + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $0$ by $0$ .

$D_{z} = - 3 (0 - 1 \cdot 4) + 3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} | + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Multiply $- 1$ by $4$ .

$D_{z} = - 3 (0 - 4) + 3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} | + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Simplify the expression.

Tap for more steps...

Subtract $4$ from $0$ .

$D_{z} = - 3 \cdot - 4 + 3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} | + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Multiply $- 3$ by $- 4$ .

$D_{z} = 12 + 3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} | + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

The determinant of $3 | \begin{array}{c} 1 & - 2 \\ 1 & 0 \end{array} |$ is $6$ .

Tap for more steps...

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

$D_{z} = 12 + 3 ((1) (0) - 1 \cdot - 2) + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $0$ by $1$ .

$D_{z} = 12 + 3 (0 - 1 \cdot - 2) + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Multiply $- 1$ by $- 2$ .

$D_{z} = 12 + 3 (0 + 2) + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Simplify the expression.

Tap for more steps...

Add $0$ and $2$ .

$D_{z} = 12 + 3 \cdot 2 + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Multiply $3$ by $2$ .

$D_{z} = 12 + 6 + 0 | \begin{array}{c} 1 & - 2 \\ 0 & 4 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{z} = 12 + 6 + 0$

Add $12$ and $6$ .

$D_{z} = 18 + 0$

Add $18$ and $0$ .

$D_{z} = 18$

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{18}{- 9}$ .

Tap for more steps...

Remove the parentheses from the numerator.

$x = \frac{18}{- 9}$

Divide $18$ by $- 9$ .

$x = - 2$

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ . In this case, $y = \frac{90}{- 9}$ .

Tap for more steps...

Remove the parentheses from the numerator.

$y = \frac{90}{- 9}$

Divide $90$ by $- 9$ .

$y = - 10$

Find the value of $z$ by Cramer's Rule, which states that $z = \frac{D_{z}}{D}$ . In this case, $z = \frac{18}{- 9}$ .

Tap for more steps...

Remove the parentheses from the numerator.

$z = \frac{18}{- 9}$

Divide $18$ by $- 9$ .

$z = - 2$

The solution to the system of equations using Cramer's Rule.

$x = - 2, y = - 10, z = - 2$

Enter YOUR Problem