Linear Algebra Examples

$y = \frac{4}{x^{2} + 2 x - 3}$

Step 1

Set the denominator in

$\frac{4}{x^{2} + 2 x - 3}$ equal to

$0$ to find where the expression is undefined.

$x^{2} + 2 x - 3 = 0$

Step 2

Solve for

$x$ .

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Step 2.1

Factor

$x^{2} + 2 x - 3$ using the AC method.

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Step 2.1.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$- 3$ and whose sum is

$2$ .

$- 1, 3$

Step 2.1.2

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

Step 2.2

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$x - 1 = 0$

$x + 3 = 0$

Step 2.3

Set

$x - 1$ equal to

$0$ and solve for

$x$ .

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Step 2.3.1

Set

$x - 1$ equal to

$0$ .

$x - 1 = 0$

Step 2.3.2

Add

$1$ to both sides of the equation.

$x = 1$

Step 2.4

Set

$x + 3$ equal to

$0$ and solve for

$x$ .

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Step 2.4.1

Set

$x + 3$ equal to

$0$ .

$x + 3 = 0$

Step 2.4.2

Subtract

$3$ from both sides of the equation.

$x = - 3$

Step 2.5

The final solution is all the values that make

$(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Step 3

The domain is all values of

$x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3) \cup (- 3, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1, - 3}$

Step 4