Linear Algebra Examples

$5 x - y + 1 = 0$

Step 1

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

Tap for more steps...

Step 1.1

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

$- y + 1 = - 5 x$

Step 1.2

Subtract $1$ from both sides of the equation.

$- y = - 5 x - 1$

Step 2

Divide each term in $- y = - 5 x - 1$ by $- 1$ and simplify.

Tap for more steps...

Step 2.1

Divide each term in $- y = - 5 x - 1$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 5 x}{- 1} + \frac{- 1}{- 1}$

Step 2.2

Simplify the left side.

Tap for more steps...

Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 5 x}{- 1} + \frac{- 1}{- 1}$

Step 2.2.2

Divide $y$ by $1$ .

$y = \frac{- 5 x}{- 1} + \frac{- 1}{- 1}$

Step 2.3

Simplify the right side.

Tap for more steps...

Step 2.3.1

Simplify each term.

Tap for more steps...

Step 2.3.1.1

Move the negative one from the denominator of $\frac{- 5 x}{- 1}$ .

$y = - 1 \cdot (- 5 x) + \frac{- 1}{- 1}$

Step 2.3.1.2

Rewrite $- 1 \cdot (- 5 x)$ as $- (- 5 x)$ .

$y = - (- 5 x) + \frac{- 1}{- 1}$

Step 2.3.1.3

Multiply $- 5$ by $- 1$ .

$y = 5 x + \frac{- 1}{- 1}$

Step 2.3.1.4

Divide $- 1$ by $- 1$ .

$y = 5 x + 1$

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:

${x | x \in ℝ}$

Step 4