Finite Math Examples

$3 z + 3 x + 3 y = 190$ , $x + 3 = y$ , $z = y - 4 x + 1$

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

$3 z + 3 x + 3 y = 190$

$x + 3 - y = 0$

$z = y - 4 x + 1$

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

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

$3 z + 3 x + 3 y = 190$

$x + 3 - y = 0$

$z - y = - 4 x + 1$

Move $- 4 x$ to the left side of the equation because it contains a variable.

$3 z + 3 x + 3 y = 190$

$x + 3 - y = 0$

$z - y + 4 x = 1$

$3 z + 3 x + 3 y = 190$

$x + 3 - y = 0$

$z - y + 4 x = 1$

Move $3$ to the right side of the equation because it does not contain a variable.

$3 z + 3 x + 3 y = 190$

$x - y = - 3$

$z - y + 4 x = 1$

Represent the system of equations in matrix format.

$[\begin{array}{c} 3 & 3 & 3 \\ 1 & - 1 & 0 \\ 4 & - 1 & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 190 \\ - 3 \\ 1 \end{array}]$

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

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Set up the determinant by breaking it into smaller components.

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

The determinant of $3 | \begin{array}{c} 1 & - 1 \\ 4 & - 1 \end{array} |$ is $9$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Simplify the expression.

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Add $- 1$ and $4$ .

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

Multiply $3$ by $3$ .

$D = 9 + 0 | \begin{array}{c} 3 & 3 \\ 4 & - 1 \end{array} | + 1 | \begin{array}{c} 3 & 3 \\ 1 & - 1 \end{array} |$

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

$D = 9 + 0 + 1 | \begin{array}{c} 3 & 3 \\ 1 & - 1 \end{array} |$

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

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

$D = 9 + 0 (3) (- 1) - 1 \cdot 3$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 1$ .

$D = 9 + 0 - 3 - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$D = 9 + 0 - 3 - 3$

Subtract $3$ from $- 3$ .

$D = 9 + 0 - 6$

Add $9$ and $0$ .

$D = 9 - 6$

Subtract $6$ from $9$ .

$D = 3$

The determinant of $[\begin{array}{c} 190 & 3 & 3 \\ - 3 & - 1 & 0 \\ 1 & - 1 & 1 \end{array}]$ is $- 169$ .

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Set up the determinant by breaking it into smaller components.

$D_{x} = 3 | \begin{array}{c} - 3 & - 1 \\ 1 & - 1 \end{array} | + 0 | \begin{array}{c} 190 & 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

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

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

$D_{x} = 3 ((- 3) (- 1) - 1 \cdot - 1) + 0 | \begin{array}{c} 190 & 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 3$ by $- 1$ .

$D_{x} = 3 (3 - 1 \cdot - 1) + 0 | \begin{array}{c} 190 & 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

Multiply $- 1$ by $- 1$ .

$D_{x} = 3 (3 + 1) + 0 | \begin{array}{c} 190 & 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

Simplify the expression.

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Add $3$ and $1$ .

$D_{x} = 3 \cdot 4 + 0 | \begin{array}{c} 190 & 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

Multiply $3$ by $4$ .

$D_{x} = 12 + 0 | \begin{array}{c} 190 & 3 \\ 1 & - 1 \end{array} | + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

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

$D_{x} = 12 + 0 + 1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$

The determinant of $1 | \begin{array}{c} 190 & 3 \\ - 3 & - 1 \end{array} |$ is $- 181$ .

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

$D_{x} = 12 + 0 (190) (- 1) + 3 \cdot 3$

Simplify the determinant.

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Simplify each term.

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Multiply $190$ by $- 1$ .

$D_{x} = 12 + 0 - 190 + 3 \cdot 3$

Multiply $3$ by $3$ .

$D_{x} = 12 + 0 - 190 + 9$

Add $- 190$ and $9$ .

$D_{x} = 12 + 0 - 181$

Add $12$ and $0$ .

$D_{x} = 12 - 181$

Subtract $181$ from $12$ .

$D_{x} = - 169$

The determinant of $[\begin{array}{c} 3 & 190 & 3 \\ 1 & - 3 & 0 \\ 4 & 1 & 1 \end{array}]$ is $- 160$ .

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Set up the determinant by breaking it into smaller components.

$D_{y} = 3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} | + 0 | \begin{array}{c} 3 & 190 \\ 4 & 1 \end{array} | + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

The determinant of $3 | \begin{array}{c} 1 & - 3 \\ 4 & 1 \end{array} |$ is $39$ .

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

$D_{y} = 3 ((1) (1) - 4 \cdot - 3) + 0 | \begin{array}{c} 3 & 190 \\ 4 & 1 \end{array} | + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $1$ by $1$ .

$D_{y} = 3 (1 - 4 \cdot - 3) + 0 | \begin{array}{c} 3 & 190 \\ 4 & 1 \end{array} | + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

Multiply $- 4$ by $- 3$ .

$D_{y} = 3 (1 + 12) + 0 | \begin{array}{c} 3 & 190 \\ 4 & 1 \end{array} | + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

Simplify the expression.

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Add $1$ and $12$ .

$D_{y} = 3 \cdot 13 + 0 | \begin{array}{c} 3 & 190 \\ 4 & 1 \end{array} | + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

Multiply $3$ by $13$ .

$D_{y} = 39 + 0 | \begin{array}{c} 3 & 190 \\ 4 & 1 \end{array} | + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

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

$D_{y} = 39 + 0 + 1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$

The determinant of $1 | \begin{array}{c} 3 & 190 \\ 1 & - 3 \end{array} |$ is $- 199$ .

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

$D_{y} = 39 + 0 (3) (- 3) - 1 \cdot 190$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 3$ .

$D_{y} = 39 + 0 - 9 - 1 \cdot 190$

Multiply $- 1$ by $190$ .

$D_{y} = 39 + 0 - 9 - 190$

Subtract $190$ from $- 9$ .

$D_{y} = 39 + 0 - 199$

Add $39$ and $0$ .

$D_{y} = 39 - 199$

Subtract $199$ from $39$ .

$D_{y} = - 160$

The determinant of $[\begin{array}{c} 3 & 3 & 190 \\ 1 & - 1 & - 3 \\ 4 & - 1 & 1 \end{array}]$ is $519$ .

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Set up the determinant by breaking it into smaller components.

$D_{z} = 3 | \begin{array}{c} - 1 & - 3 \\ - 1 & 1 \end{array} | - 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} | + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

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

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

$D_{z} = 3 ((- 1) (1) + 1 \cdot - 3) - 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} | + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $- 1$ by $1$ .

$D_{z} = 3 (- 1 + 1 \cdot - 3) - 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} | + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Multiply $- 3$ by $1$ .

$D_{z} = 3 (- 1 - 3) - 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} | + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Simplify the expression.

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Subtract $3$ from $- 1$ .

$D_{z} = 3 \cdot - 4 - 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} | + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Multiply $3$ by $- 4$ .

$D_{z} = - 12 - 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} | + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

The determinant of $- 1 | \begin{array}{c} 3 & 190 \\ - 1 & 1 \end{array} |$ is $- 193$ .

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

$D_{z} = - 12 - 1 ((3) (1) + 1 \cdot 190) + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $1$ .

$D_{z} = - 12 - 1 (3 + 1 \cdot 190) + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Multiply $190$ by $1$ .

$D_{z} = - 12 - 1 (3 + 190) + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Simplify the expression.

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Add $3$ and $190$ .

$D_{z} = - 12 - 1 \cdot 193 + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

Multiply $- 1$ by $193$ .

$D_{z} = - 12 - 193 + 4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$

The determinant of $4 | \begin{array}{c} 3 & 190 \\ - 1 & - 3 \end{array} |$ is $724$ .

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

$D_{z} = - 12 - 193 + 4 ((3) (- 3) + 1 \cdot 190)$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $- 3$ .

$D_{z} = - 12 - 193 + 4 (- 9 + 1 \cdot 190)$

Multiply $190$ by $1$ .

$D_{z} = - 12 - 193 + 4 (- 9 + 190)$

Simplify the expression.

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Add $- 9$ and $190$ .

$D_{z} = - 12 - 193 + 4 \cdot 181$

Multiply $4$ by $181$ .

$D_{z} = - 12 - 193 + 724$

Subtract $193$ from $- 12$ .

$D_{z} = - 205 + 724$

Add $- 205$ and $724$ .

$D_{z} = 519$

Remove the extra parentheses from the expression $\frac{- 169}{3}$ .

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Remove the parentheses from the numerator.

$x = - \frac{169}{3}$

Remove the parentheses from the denominator.

$x = - \frac{169}{3}$

Remove the extra parentheses from the expression $\frac{- 160}{3}$ .

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Remove the parentheses from the numerator.

$y = - \frac{160}{3}$

Remove the parentheses from the denominator.

$y = - \frac{160}{3}$

Find the value of $z$ by Cramer's Rule, which states that $z = \frac{D_{z}}{D}$ . In this case, $z = \frac{519}{3}$ .

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Remove the extra parentheses from the expression $\frac{519}{3}$ .

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Remove the parentheses from the numerator.

$z = \frac{519}{3}$

Remove the parentheses from the denominator.

$z = \frac{519}{3}$

Divide $519$ by $3$ .

$z = 173$

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

$x = - \frac{169}{3}, y = - \frac{160}{3}, z = 173$

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