QUESTION IMAGE
Question
- \\(\frac{3}{p+4} + \frac{2p+32}{p^2 - 16}\\)\
- \\(\frac{n^2 + 7n - 18}{1} - \frac{1}{n^2 - 5n + 6} \cdot \frac{1}{n - 3}\\)\
- \\(\frac{2}{y - 7} + \frac{1}{y - 4}\\)\
- \\(\frac{r}{r^2 - 9} - \frac{1}{r^2 + 8r + 15}\\)\
directions: simplify each complex fraction.\
- \\(\frac{4 - \frac{1}{k}}{\frac{8}{k} - \frac{2}{k^2}}\\)\
- \\(3 + \frac{\frac{v}{2}}{\frac{v}{4} - \frac{9}{v}}\\)\
- \\(\frac{2a}{9b} - \frac{b}{2a} \div \frac{a^3 + 1}{3b^3 + 2b}\\)\
- \\(x^2 + \frac{1}{x} \div \frac{(x + 1)^2}{4x}\\)\
- write an expression to represent the area of the rectangle below in simplest form. one side: \\(\frac{2x^2 - 32x}{x + 4}\\), the other: \\(\frac{1 - \frac{4}{x^2}}{4(x - 4)^2}\\)
Let's solve problem 10: $\frac{3}{p + 4}+\frac{2p + 32}{p^{2}-16}$
Step 1: Factor the denominator
Notice that $p^{2}-16$ is a difference of squares, so $p^{2}-16=(p + 4)(p - 4)$
The expression becomes $\frac{3}{p + 4}+\frac{2p + 32}{(p + 4)(p - 4)}$
Step 2: Find a common denominator
The common denominator for the two fractions is $(p + 4)(p - 4)$
Rewrite the first fraction with the common denominator: $\frac{3(p - 4)}{(p + 4)(p - 4)}+\frac{2p + 32}{(p + 4)(p - 4)}$
Step 3: Expand the numerator of the first fraction
$3(p - 4)=3p-12$
So the expression is $\frac{3p - 12}{(p + 4)(p - 4)}+\frac{2p + 32}{(p + 4)(p - 4)}$
Step 4: Add the numerators
$(3p - 12)+(2p + 32)=3p-12 + 2p+32=5p + 20$
The fraction becomes $\frac{5p + 20}{(p + 4)(p - 4)}$
Step 5: Factor the numerator
$5p + 20 = 5(p + 4)$
So we have $\frac{5(p + 4)}{(p + 4)(p - 4)}$
Step 6: Cancel out the common factor
Cancel out $(p + 4)$ from the numerator and the denominator (assuming $p
eq - 4$)
We get $\frac{5}{p - 4}$
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$\frac{5}{p - 4}$