Finite Math Examples

$x + y - z = 12$ , $x + 2 y + z = 32$ , $x + y = 20$

Step 1

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

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Step 1.1

Subtract $y$ from both sides of the equation.

$x - z = 12 - y$

$x + 2 y + z = 32$

$x + y = 20$

Step 1.2

Add $z$ to both sides of the equation.

$x = 12 - y + z$

$x + 2 y + z = 32$

$x + y = 20$

$x = 12 - y + z$

$x + 2 y + z = 32$

$x + y = 20$

Step 2

Replace all occurrences of $x$ with $12 - y + z$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x + 2 y + z = 32$ with $12 - y + z$ .

$(12 - y + z) + 2 y + z = 32$

$x = 12 - y + z$

$x + y = 20$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $(12 - y + z) + 2 y + z$ .

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Step 2.2.1.1

Add $- y$ and $2 y$ .

$12 + y + z + z = 32$

$x = 12 - y + z$

$x + y = 20$

Step 2.2.1.2

Add $z$ and $z$ .

$12 + y + 2 z = 32$

$x = 12 - y + z$

$x + y = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

$x + y = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

$x + y = 20$

Step 2.3

Replace all occurrences of $x$ in $x + y = 20$ with $12 - y + z$ .

$(12 - y + z) + y = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

Step 2.4

Simplify the left side.

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Step 2.4.1

Combine the opposite terms in $(12 - y + z) + y$ .

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Step 2.4.1.1

Add $- y$ and $y$ .

$12 + z + 0 = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

Step 2.4.1.2

Add $12 + z$ and $0$ .

$12 + z = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

$12 + z = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

$12 + z = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

$12 + z = 20$

$12 + y + 2 z = 32$

$x = 12 - y + z$

Step 3

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

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Step 3.1

Subtract $12$ from both sides of the equation.

$z = 20 - 12$

$12 + y + 2 z = 32$

$x = 12 - y + z$

Step 3.2

Subtract $12$ from $20$ .

$z = 8$

$12 + y + 2 z = 32$

$x = 12 - y + z$

$z = 8$

$12 + y + 2 z = 32$

$x = 12 - y + z$

Step 4

Replace all occurrences of $z$ with $8$ in each equation.

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Step 4.1

Replace all occurrences of $z$ in $12 + y + 2 z = 32$ with $8$ .

$12 + y + 2 (8) = 32$

$z = 8$

$x = 12 - y + z$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $12 + y + 2 (8)$ .

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Step 4.2.1.1

Multiply $2$ by $8$ .

$12 + y + 16 = 32$

$z = 8$

$x = 12 - y + z$

Step 4.2.1.2

Add $12$ and $16$ .

$y + 28 = 32$

$z = 8$

$x = 12 - y + z$

$y + 28 = 32$

$z = 8$

$x = 12 - y + z$

$y + 28 = 32$

$z = 8$

$x = 12 - y + z$

Step 4.3

Replace all occurrences of $z$ in $x = 12 - y + z$ with $8$ .

$x = 12 - y + 8$

$y + 28 = 32$

$z = 8$

Step 4.4

Simplify $x = 12 - y + 8$ .

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Step 4.4.1

Simplify the left side.

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Step 4.4.1.1

Remove parentheses.

$x = 12 - y + 8$

$y + 28 = 32$

$z = 8$

$x = 12 - y + 8$

$y + 28 = 32$

$z = 8$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Add $12$ and $8$ .

$x = - y + 20$

$y + 28 = 32$

$z = 8$

$x = - y + 20$

$y + 28 = 32$

$z = 8$

$x = - y + 20$

$y + 28 = 32$

$z = 8$

$x = - y + 20$

$y + 28 = 32$

$z = 8$

Step 5

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

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Step 5.1

Subtract $28$ from both sides of the equation.

$y = 32 - 28$

$x = - y + 20$

$z = 8$

Step 5.2

Subtract $28$ from $32$ .

$y = 4$

$x = - y + 20$

$z = 8$

$y = 4$

$x = - y + 20$

$z = 8$

Step 6

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

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Step 6.1

Replace all occurrences of $y$ in $x = - y + 20$ with $4$ .

$x = - (4) + 20$

$y = 4$

$z = 8$

Step 6.2

Simplify the right side.

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Step 6.2.1

Simplify $- (4) + 20$ .

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Step 6.2.1.1

Multiply $- 1$ by $4$ .

$x = - 4 + 20$

$y = 4$

$z = 8$

Step 6.2.1.2

Add $- 4$ and $20$ .

$x = 16$

$y = 4$

$z = 8$

$x = 16$

$y = 4$

$z = 8$

$x = 16$

$y = 4$

$z = 8$

$x = 16$

$y = 4$

$z = 8$

Step 7

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

$(16, 4, 8)$

Step 8

The result can be shown in multiple forms.

Point Form:

$(16, 4, 8)$

Equation Form:

$x = 16, y = 4, z = 8$