Linear Algebra Examples

$4 x - 2 y - 8 z = - 8$

Step 1

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

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Step 1.1

Subtract $4 x$ from both sides of the equation.

$- 2 y - 8 z = - 8 - 4 x$

Step 1.2

Add $8 z$ to both sides of the equation.

$- 2 y = - 8 - 4 x + 8 z$

Step 2

Divide each term in $- 2 y = - 8 - 4 x + 8 z$ by $- 2$ and simplify.

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Step 2.1

Divide each term in $- 2 y = - 8 - 4 x + 8 z$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{- 8}{- 2} + \frac{- 4 x}{- 2} + \frac{8 z}{- 2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $- 2$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{- 8}{- 2} + \frac{- 4 x}{- 2} + \frac{8 z}{- 2}$

Step 2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 8}{- 2} + \frac{- 4 x}{- 2} + \frac{8 z}{- 2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Divide $- 8$ by $- 2$ .

$y = 4 + \frac{- 4 x}{- 2} + \frac{8 z}{- 2}$

Step 2.3.1.2

Cancel the common factor of $- 4$ and $- 2$ .

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Step 2.3.1.2.1

Factor $- 2$ out of $- 4 x$ .

$y = 4 + \frac{- 2 (2 x)}{- 2} + \frac{8 z}{- 2}$

Step 2.3.1.2.2

Cancel the common factors.

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Step 2.3.1.2.2.1

Factor $- 2$ out of $- 2$ .

$y = 4 + \frac{- 2 (2 x)}{- 2 (1)} + \frac{8 z}{- 2}$

Step 2.3.1.2.2.2

Cancel the common factor.

$y = 4 + \frac{- 2 (2 x)}{- 2 \cdot 1} + \frac{8 z}{- 2}$

Step 2.3.1.2.2.3

Rewrite the expression.

$y = 4 + \frac{2 x}{1} + \frac{8 z}{- 2}$

Step 2.3.1.2.2.4

Divide $2 x$ by $1$ .

$y = 4 + 2 x + \frac{8 z}{- 2}$

Step 2.3.1.3

Cancel the common factor of $8$ and $- 2$ .

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Step 2.3.1.3.1

Factor $2$ out of $8 z$ .

$y = 4 + 2 x + \frac{2 (4 z)}{- 2}$

Step 2.3.1.3.2

Move the negative one from the denominator of $\frac{4 z}{- 1}$ .

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

Step 2.3.1.4

Rewrite $- 1 \cdot (4 z)$ as $- (4 z)$ .

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

Step 2.3.1.5

Multiply $4$ by $- 1$ .

$y = 4 + 2 x - 4 z$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$