Solving for $x$: $\;\frac8{x-2}-\frac{13}2=\frac3{2x-4}.$

Question

I have solved this sort of problem before, but frustratingly I have forgotten to.

The problem is:

Solve for $x$: $$\dfrac8{x-2}-\dfrac{13}2=\dfrac3{2x-4}.$$

So hence the title, How should I solve this?

Answer 1

$$\frac{8}{x-2}-\frac{13}{2}=\frac{3}{2x-4}$$

We want all the denominators to be the same, so we rewrite each fraction.

$$\frac{8*2}{2(x-2)}-\frac{13(x-2)}{2(x-2)}=\frac{3}{2x-4}$$

We multiply both sides by $2x-4$ and continue solving for $x$:

$$16-13(x-2)=3$$

$$16-13x+26=3$$

$$13x=39$$

$$x=3$$

Answer 2

First note the domain of the function: All reals such that $x\neq 2$. Then multiply both sides by $x-2$. This will give you a linear equation that you can easily solve. If the solution is not 2, then there is a solution. Otherwise, there is no solution.

Answer 3

Find the common denominator on the left. Note that when we do this, the denominator on the left hand side is equal to that of the right-hand side. So the numerators of each side must be equal.

$$\frac{8}{x-2}-\frac{13}2 = \frac{16 - 13(x - 2)}{2(x - 2)} = \frac {42 - 13x}{2x-4} = \frac 3{2x-4}$$

$$\iff 42-13x = 3$$ $$\iff 13x = 39 \iff x= 3$$

Answer 4

It is probably simplest to note at the very beginning that the lcm of the denominators is 2x - 4 and 1/(2x - 4) can never be zero.

Therefore, multiply throughout by 2x - 4 to get:

$8 \times 2 - 13 \times (x - 2) = 3$

So $13x = 26 + 16 - 3 = 39$.

This gives x = 3.