Linear Algebra Examples

x - y = - 3

$x - y = - 3$

Step 1

Subtract

x

$x$ from both sides of the equation.

- y = - 3 - x

$- y = - 3 - x$

Step 2

Divide each term in

- y = - 3 - x

$- y = - 3 - x$ by

- 1

$- 1$ and simplify.

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Step 2.1

Divide each term in

- y = - 3 - x

$- y = - 3 - x$ by

- 1

$- 1$ .

\frac{- y}{- 1} = \frac{- 3}{- 1} + \frac{- x}{- 1}

$\frac{- y}{- 1} = \frac{- 3}{- 1} + \frac{- x}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

\frac{y}{1} = \frac{- 3}{- 1} + \frac{- x}{- 1}

$\frac{y}{1} = \frac{- 3}{- 1} + \frac{- x}{- 1}$

Step 2.2.2

Divide

y

$y$ by

1

$1$ .

y = \frac{- 3}{- 1} + \frac{- x}{- 1}

$y = \frac{- 3}{- 1} + \frac{- x}{- 1}$

y = \frac{- 3}{- 1} + \frac{- x}{- 1}

$y = \frac{- 3}{- 1} + \frac{- x}{- 1}$

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

- 3

$- 3$ by

- 1

$- 1$ .

y = 3 + \frac{- x}{- 1}

$y = 3 + \frac{- x}{- 1}$

Step 2.3.1.2

Dividing two negative values results in a positive value.

y = 3 + \frac{x}{1}

$y = 3 + \frac{x}{1}$

Step 2.3.1.3

Divide

x

$x$ by

1

$1$ .

y = 3 + x

$y = 3 + x$

y = 3 + x

$y = 3 + x$

y = 3 + x

$y = 3 + x$

y = 3 + x

$y = 3 + x$

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)

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

$x - y = - 3$

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