I'm self-studying from Stroud's amazing "Engineering Mathematics" textbook, and have run into an issue in one of the final end-of-chapter exercises for in the "Linear Equations" chapter.
Here's the problem:
Writing $u$ for $\dfrac{1}{x+8y}$ and $v$ for $\dfrac{1}{8x-y}$, solve the following equations for $u$ and $v$, and hence determine the values of $x$ and $y$:
$$\frac{2}{x+8y} - \frac{1}{8x-y} = 4;$$ $$\frac{1}{x+8y} + \frac{2}{8x-y} = 7$$
So far, in my solution I have
$$2u - v = 4$$ $$u + 2v = 7$$
Multiplying the first one by 2:
$$4u - 4v = 8$$ $$u + 4v = 7$$
Subtracting, we can find the value of $u$: $$5u = 15$$ $$u = 3$$
Now we can easily find the value of $v$:
$$4v = 7 - u$$ $$4v = v$$ $$v = 4$$
So, my question is - what next?
I assume that $u = \frac{1}{x + 8y}$, instead of $u = \frac{1}{x+8}.$
I did not check your computations, but the approach remains the same. You have found that $$\begin{cases} u = 3 &= \frac{1}{x+8y}\\ v = 4 &= \frac{1}{8x-y} \end{cases}$$ This implies that $$\begin{cases} x + 8y &= \frac{1}{3}\\ 8x-y &= \frac{1}{4} \end{cases}.$$ Can you take it from here?