Question #bf791

Question

Question #bf791

1 Answer

Impact of this question

4847 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-03T03:05:26

The answer is (B) $\frac{x^{2} + 5 x + 6}{2 x + 5}$ $(x^2 + 5x + 6)/(2x+5)$

Explanation:

The idea here is that you need to take your starting eexpressions

$\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}}$ $1/(1/(x+2) + 1/(x+3))$

and do some algebraic manipulations to get it to math one of the four expressions given.

Focus on the denominator of the original fraction

$\frac{1}{x + 2} + \frac{1}{x + 3}$ $1/(x+2) + 1/(x+3)$

In order to be able to add these two fractions, you need them to have the same denominator. This means that you're going to have to multiply the first one by $1 = \frac{x + 3}{x + 3}$ $1 = (x+3)/(x+3)$ and the second one by $1 = \frac{x + 2}{x + 2}$ $1 = (x+2)/(x+2)$ to get their common denominator, $(x + 2) (x + 3)$ $(x+2)(x+3)$

$\frac{1}{x + 2} \cdot \frac{x + 3}{x + 3} + \frac{1}{x + 3} \cdot \frac{x + 2}{x + 2}$ $1/(x+2) * (x+3)/(x+3) + 1/(x+3) * (x+2)/(x+2)$

$\frac{x + 3}{(x + 2) (x + 3)} + \frac{x + 2}{(x + 2) (x + 3)}$ $(x+3)/((x+2)(x+3)) + (x+2)/((x+2)(x+3))$

Now you can add these fractions by adding their numerators

$\frac{x + 3 + x + 2}{(x + 2) (x + 3)} = \frac{2 x + 5}{(x + 2) (x + 3)}$ $( x+3 + x+2)/((x+2)(x+3)) = (2x + 5)/((x+2)(x+3))$

The original fraction can thus be written as

$\frac{1}{\frac{2 x + 5}{(x + 2) (x + 3)}}$ $1/((2x+5)/((x+2)(x+3))$

Now, you know that dividing a number by a fraction is equivalent to multiplying the number by the inverse of the fraction

$\frac{a}{\frac{b}{c}} = a \cdot \frac{c}{b}$ $color(blue)(a/(b/c) = a * c/b)$

In your case, $a = 1$ $a = 1$ , $b = (2 x + 5)$ $b = (2x+5)$ and $c = (x + 2) (x + 3)$ $c = (x+2)(x+3)$ . This means that you have

$\frac{1}{\frac{2 x + 5}{(x + 2) (x + 3)}} = 1 \cdot \frac{(x + 2) (x + 3)}{2 x + 5} = \frac{(x + 2) (x + 3)}{2 x + 5}$ $1/((2x+5)/((x+2)(x+3))) = 1 * ((x+2)(x+3))/(2x+5) = ((x+2)(x+3))/(2x+5)$

Finally, you can expand the parantheses of the numerator to get

$(x + 2) (x + 3) = x^{2} + 2 x + 3 x + 6 = x^{2} + 5 x + 6$ $(x+2)(x+3) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6$

This means that the expression will be equal to

$\frac{(x + 2) (x + 3)}{2 x + 5} = \frac{x^{2} + 5 x + 6}{2 x + 5}$ $((x+2)(x+3))/(2x+5) = (x^2 + 5x + 6)/(2x+5)$

Therefore, you have

$\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}} = \frac{x^{2} + 5 x + 6}{2 x + 5}$ $1/(1/(x+2) + 1/(x+3)) = color(green)((x^2 + 5x + 6)/(2x+5))$

Science

Math

Humanities

... and beyond

Question #bf791

1 Answer

Explanation:

Related questions

Impact of this question