Algebra Examples

$r (x) = x + 3$ ; $s (x) = 4 x + 3$

Step 1

Set up the composite result function.

$s (r (x))$

Step 2

Evaluate $s (x + 3)$ by substituting in the value of $r$ into $s$ .

$s (x + 3) = 4 (x + 3) + 3$

Step 3

Simplify each term.

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Step 3.1

Apply the distributive property.

$s (x + 3) = 4 x + 4 \cdot 3 + 3$

Step 3.2

Multiply $4$ by $3$ .

$s (x + 3) = 4 x + 12 + 3$

Step 4

Add $12$ and $3$ .

$s (x + 3) = 4 x + 15$