Finite Math Examples

$x - y + 5 z = 13$ , $y - 2 z = - 7$ , $y + 8 z = 33$

Step 1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.1

Add $y$ to both sides of the equation.

$x + 5 z = 13 + y$

$y - 2 z = - 7$

$y + 8 z = 33$

Step 1.2

Subtract $5 z$ from both sides of the equation.

$x = 13 + y - 5 z$

$y - 2 z = - 7$

$y + 8 z = 33$

$x = 13 + y - 5 z$

$y - 2 z = - 7$

$y + 8 z = 33$

Step 2

Add $2 z$ to both sides of the equation.

$y = - 7 + 2 z$

$x = 13 + y - 5 z$

$y + 8 z = 33$

Step 3

Replace all occurrences of $y$ with $- 7 + 2 z$ in each equation.

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Step 3.1

Replace all occurrences of $y$ in $x = 13 + y - 5 z$ with $- 7 + 2 z$ .

$x = 13 - 7 + 2 z - 5 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

Step 3.2

Simplify $x = 13 - 7 + 2 z - 5 z$ .

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Remove parentheses.

$x = 13 - 7 + 2 z - 5 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

$x = 13 - 7 + 2 z - 5 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $13 - 7 + 2 z - 5 z$ .

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Step 3.2.2.1.1

Subtract $7$ from $13$ .

$x = 6 + 2 z - 5 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

Step 3.2.2.1.2

Subtract $5 z$ from $2 z$ .

$x = 6 - 3 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$y + 8 z = 33$

Step 3.3

Replace all occurrences of $y$ in $y + 8 z = 33$ with $- 7 + 2 z$ .

$(- 7 + 2 z) + 8 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 3.4

Simplify the left side.

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Step 3.4.1

Add $2 z$ and $8 z$ .

$- 7 + 10 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$- 7 + 10 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$- 7 + 10 z = 33$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 4

Solve for $z$ in $- 7 + 10 z = 33$ .

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Step 4.1

Move all terms not containing $z$ to the right side of the equation.

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Step 4.1.1

Add $7$ to both sides of the equation.

$10 z = 33 + 7$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 4.1.2

Add $33$ and $7$ .

$10 z = 40$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$10 z = 40$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 4.2

Divide each term in $10 z = 40$ by $10$ and simplify.

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Step 4.2.1

Divide each term in $10 z = 40$ by $10$ .

$\frac{10 z}{10} = \frac{40}{10}$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $10$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{10 z}{10} = \frac{40}{10}$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 4.2.2.1.2

Divide $z$ by $1$ .

$z = \frac{40}{10}$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$z = \frac{40}{10}$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$z = \frac{40}{10}$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $40$ by $10$ .

$z = 4$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$z = 4$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$z = 4$

$x = 6 - 3 z$

$y = - 7 + 2 z$

$z = 4$

$x = 6 - 3 z$

$y = - 7 + 2 z$

Step 5

Replace all occurrences of $z$ with $4$ in each equation.

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Step 5.1

Replace all occurrences of $z$ in $x = 6 - 3 z$ with $4$ .

$x = 6 - 3 \cdot 4$

$z = 4$

$y = - 7 + 2 z$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $6 - 3 \cdot 4$ .

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Step 5.2.1.1

Multiply $- 3$ by $4$ .

$x = 6 - 12$

$z = 4$

$y = - 7 + 2 z$

Step 5.2.1.2

Subtract $12$ from $6$ .

$x = - 6$

$z = 4$

$y = - 7 + 2 z$

$x = - 6$

$z = 4$

$y = - 7 + 2 z$

$x = - 6$

$z = 4$

$y = - 7 + 2 z$

Step 5.3

Replace all occurrences of $z$ in $y = - 7 + 2 z$ with $4$ .

$y = - 7 + 2 (4)$

$x = - 6$

$z = 4$

Step 5.4

Simplify the right side.

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Step 5.4.1

Simplify $- 7 + 2 (4)$ .

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Step 5.4.1.1

Multiply $2$ by $4$ .

$y = - 7 + 8$

$x = - 6$

$z = 4$

Step 5.4.1.2

Add $- 7$ and $8$ .

$y = 1$

$x = - 6$

$z = 4$

$y = 1$

$x = - 6$

$z = 4$

$y = 1$

$x = - 6$

$z = 4$

$y = 1$

$x = - 6$

$z = 4$

Step 6

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(- 6, 1, 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 6, 1, 4)$

Equation Form:

$x = - 6, y = 1, z = 4$