Precalculus Examples

$y - 3 z = 0$ , $3 x + 2 y = 12$ , $- x - 2 y + 4 z = - 4$

Step 1

Add $3 z$ to both sides of the equation.

$y = 3 z$

$3 x + 2 y = 12$

$- x - 2 y + 4 z = - 4$

Step 2

Replace all occurrences of $y$ with $3 z$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $3 x + 2 y = 12$ with $3 z$ .

$3 x + 2 (3 z) = 12$

$y = 3 z$

$- x - 2 y + 4 z = - 4$

Step 2.2

Simplify the left side.

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Step 2.2.1

Multiply $3$ by $2$ .

$3 x + 6 z = 12$

$y = 3 z$

$- x - 2 y + 4 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

$- x - 2 y + 4 z = - 4$

Step 2.3

Replace all occurrences of $y$ in $- x - 2 y + 4 z = - 4$ with $3 z$ .

$- x - 2 (3 z) + 4 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

Step 2.4

Simplify the left side.

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Step 2.4.1

Simplify $- x - 2 (3 z) + 4 z$ .

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Step 2.4.1.1

Multiply $3$ by $- 2$ .

$- x - 6 z + 4 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

Step 2.4.1.2

Add $- 6 z$ and $4 z$ .

$- x - 2 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

$- x - 2 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

$- x - 2 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

$- x - 2 z = - 4$

$3 x + 6 z = 12$

$y = 3 z$

Step 3

Solve for $x$ in $- x - 2 z = - 4$ .

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Step 3.1

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

$- x = - 4 + 2 z$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2

Divide each term in $- x = - 4 + 2 z$ by $- 1$ and simplify.

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Step 3.2.1

Divide each term in $- x = - 4 + 2 z$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1} + \frac{2 z}{- 1}$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1} + \frac{2 z}{- 1}$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1} + \frac{2 z}{- 1}$

$3 x + 6 z = 12$

$y = 3 z$

$x = \frac{- 4}{- 1} + \frac{2 z}{- 1}$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Divide $- 4$ by $- 1$ .

$x = 4 + \frac{2 z}{- 1}$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2.3.1.2

Move the negative one from the denominator of $\frac{2 z}{- 1}$ .

$x = 4 - 1 \cdot (2 z)$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2.3.1.3

Rewrite $- 1 \cdot (2 z)$ as $- (2 z)$ .

$x = 4 - (2 z)$

$3 x + 6 z = 12$

$y = 3 z$

Step 3.2.3.1.4

Multiply $2$ by $- 1$ .

$x = 4 - 2 z$

$3 x + 6 z = 12$

$y = 3 z$

$x = 4 - 2 z$

$3 x + 6 z = 12$

$y = 3 z$

$x = 4 - 2 z$

$3 x + 6 z = 12$

$y = 3 z$

$x = 4 - 2 z$

$3 x + 6 z = 12$

$y = 3 z$

$x = 4 - 2 z$

$3 x + 6 z = 12$

$y = 3 z$

Step 4

Replace all occurrences of $x$ with $4 - 2 z$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $3 x + 6 z = 12$ with $4 - 2 z$ .

$3 (4 - 2 z) + 6 z = 12$

$x = 4 - 2 z$

$y = 3 z$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $3 (4 - 2 z) + 6 z$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Apply the distributive property.

$3 \cdot 4 + 3 (- 2 z) + 6 z = 12$

$x = 4 - 2 z$

$y = 3 z$

Step 4.2.1.1.2

Multiply $3$ by $4$ .

$12 + 3 (- 2 z) + 6 z = 12$

$x = 4 - 2 z$

$y = 3 z$

Step 4.2.1.1.3

Multiply $- 2$ by $3$ .

$12 - 6 z + 6 z = 12$

$x = 4 - 2 z$

$y = 3 z$

$12 - 6 z + 6 z = 12$

$x = 4 - 2 z$

$y = 3 z$

Step 4.2.1.2

Combine the opposite terms in $12 - 6 z + 6 z$ .

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Step 4.2.1.2.1

Add $- 6 z$ and $6 z$ .

$12 + 0 = 12$

$x = 4 - 2 z$

$y = 3 z$

Step 4.2.1.2.2

Add $12$ and $0$ .

$12 = 12$

$x = 4 - 2 z$

$y = 3 z$

$12 = 12$

$x = 4 - 2 z$

$y = 3 z$

$12 = 12$

$x = 4 - 2 z$

$y = 3 z$

$12 = 12$

$x = 4 - 2 z$

$y = 3 z$

$12 = 12$

$x = 4 - 2 z$

$y = 3 z$

Step 5

Remove any equations from the system that are always true.

$x = 4 - 2 z$

$y = 3 z$