Algebra Examples

$f (x) = \frac{1}{8 x + 3}$ $g (x) = 12 - 5 x$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (12 - 5 x)$ by substituting in the value of $g$ into $f$ .

$f (12 - 5 x) = \frac{1}{8 (12 - 5 x) + 3}$

Step 3

Simplify the denominator.

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Step 3.1

Apply the distributive property.

$f (12 - 5 x) = \frac{1}{8 \cdot 12 + 8 (- 5 x) + 3}$

Step 3.2

Multiply $8$ by $12$ .

$f (12 - 5 x) = \frac{1}{96 + 8 (- 5 x) + 3}$

Step 3.3

Multiply $- 5$ by $8$ .

$f (12 - 5 x) = \frac{1}{96 - 40 x + 3}$

Step 3.4

Add $96$ and $3$ .

$f (12 - 5 x) = \frac{1}{- 40 x + 99}$

Step 4

Simplify with factoring out.

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Step 4.1

Factor $- 1$ out of $- 40 x$ .

$f (12 - 5 x) = \frac{1}{- (40 x) + 99}$

Step 4.2

Rewrite $99$ as $- 1 (- 99)$ .

$f (12 - 5 x) = \frac{1}{- (40 x) - 1 \cdot - 99}$

Step 4.3

Factor $- 1$ out of $- (40 x) - 1 (- 99)$ .

$f (12 - 5 x) = \frac{1}{- (40 x - 99)}$

Step 4.4

Rewrite negatives.

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Step 4.4.1

Rewrite $- (40 x - 99)$ as $- 1 (40 x - 99)$ .

$f (12 - 5 x) = \frac{1}{- 1 (40 x - 99)}$

Step 4.4.2

Move the negative in front of the fraction.

$f (12 - 5 x) = - \frac{1}{40 x - 99}$