Linear Algebra Examples

$x^{2} y + x y = 1$

Step 1

Factor $x y$ out of $x^{2} y + x y$ .

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Step 1.1

Factor $x y$ out of $x^{2} y$ .

$x y (x) + x y = 1$

Step 1.2

Factor $x y$ out of $x y$ .

$x y (x) + x y (1) = 1$

Step 1.3

Factor $x y$ out of $x y (x) + x y (1)$ .

$x y (x + 1) = 1$

Step 2

Divide each term in $x y (x + 1) = 1$ by $x (x + 1)$ and simplify.

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Step 2.1

Divide each term in $x y (x + 1) = 1$ by $x (x + 1)$ .

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $x$ .

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Step 2.2.1.1

Cancel the common factor.

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

Step 2.2.1.2

Rewrite the expression.

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

Step 2.2.2

Cancel the common factor of $x + 1$ .

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Step 2.2.2.1

Cancel the common factor.

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

Step 2.2.2.2

Divide $y$ by $1$ .

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

Step 3

Set the denominator in $\frac{1}{x (x + 1)}$ equal to $0$ to find where the expression is undefined.

$x (x + 1) = 0$

Step 4

Solve for $x$ .

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Step 4.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 1 = 0$

Step 4.2

Set $x$ equal to $0$ .

$x = 0$

Step 4.3

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 4.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.4

The final solution is all the values that make $x (x + 1) = 0$ true.

$x = 0, - 1$

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 1}$

Step 6