# Algebraic Fractions: A reader asks

Oo, a question to answer! This one’s from Deborah.

How would you solve: $\frac{1}{x-2} + \frac{1}{x} = \frac{3}{4}$ and $\frac{1}{x}+\frac{1}{2x+1}=\frac{7}{10}$?

When I substitute my answer back in, I can’t get it to work out!

Forgive me if my working is a bit pedestrian — I figure it’s better to err on the side of too helpful rather than not helpful enough.

To solve $\frac{1}{x-2} + \frac{1}{x} = \frac{3}{4}$, I’d combine the left hand side into a single fraction, multiplying the first one by $\frac{x}{x}$ and the second by $\frac{(x-2)}{(x-2)}$ to get:

$\frac{x}{x(x-2)} + \frac{(x-2)}{x(x-2)} = \frac{2x-2}{x(x-2)} = \frac{3}{4}$

Cross-multiplying:

$4(2x - 2) = 3x(x-2) \\ 8x - 8 = 3x^2 - 6x\\ 0 = 3x^2 - 14x +8 \\ 0 = (3x-2)(x-4) $

Then either $x = \frac{2}{3}$ or $x = 4$. Substituting either value back in as $x$ on the left hand side does indeed give $\frac{3}{4}$.

The second one is similar:

$\frac{1}{x} + \frac{1}{2x+1} = \frac{7}{10} \\ \frac{2x+1}{x(2x+1)} + \frac{x}{x(2x+1)} = \frac{7}{10} \\ \frac{3x+1}{x(2x+1)} = \frac{7}{10} \\ 10(3x+1) = 7x(2x+1) \\ 30x + 10 = 14x^2 + 7x \\ 0 = 14x^2 - 23x - 10 \\ 0 = (14x+5)(x-2)$

(Yikes!) So $x = 2$ or $x = -\frac{5}{14}$. It’s clear enough that 2 works, but I’ll need to work it out for $-\frac{5}{14}$:

$\frac{1}{-5/14} + \frac{1}{-5/7 + 1} = -\frac{14}{5} + \frac{7}{2} = -\frac{28}{10} + \frac{35}{10} = \frac{7}{10}$

Whew!

*If you’d like me to answer any of your questions, you can improve your chances by showing me that you’ve made a decent stab at it first, like Deborah did.*