A fraction $\tfrac{x}{y}$ is such that when I add $8$ to $x$ and $12$ to $y$, the value of fraction unchanged

Question

I am struggling with this algebraic question where I don't know how to use the information given to work out the values for $x$ and $y$.

Question: A fraction $\tfrac{x}{y}$ is such that when I add $8$ to $x$ and $12$ to $y$, the value of fraction unchanged. What could be the numbers $x$ and $y$?

My attempt: \begin{align} & \dfrac{x}{y} = \dfrac{x + 8}{y - 12} \\ \implies & xy + 8y = xy - 12 x \end{align}

What do I do next?

Thank You and Help is Appreciated

Answer 1

if i understood correct, you will get by cross multiplication $$x(y-12)=(x+8)y$$ expanding we obtain $$-12x=8y$$ it was $$\frac{x}{y}=\frac{x+8}{y-12}$$? then we get $$-12x=8y$$ and nothing else

Answer 2

First of all, the equation by you is wrong (according to the statement preceding it). The correct one gives $12x=8y$ implying $3x=2y$

Now since $x$ and $y$ should be integers, the possible values are $(2,3)$, $(4,6)$ etc. And in general, $(2k,3k)$ All of these give you the same value of original fraction which is $2/3$.

Answer 3

$\begin{align} & \dfrac{x}{y} = \dfrac{x + 8}{y + 12} \\ \implies & xy + 8y = xy +12 x \end{align}$

(I corrected your expression. It should have been +12; not - 12. But that doesn't effect how you would solve this.)

So continue:

$8y = 12x$

$x = \frac 23 y$

Which means $y$ can be any value (except $0$ or $-12$) and $x$ will be $\frac 23y$.

Notice: $\frac {\frac 23 y + 8}{ y + 12} = \frac {2y + 24}{3y + 36} = \frac {2(y+12)}{3(y+12} = \frac 23 = \frac {\frac 23 y}{y}$.

.... or to put it another way....

So $x = 2a$ and $y=3a$ for any $a\ne 0, - 4$.

And $\frac {2a + 8}{3a+12} = \frac {2(a+4)}{3(a +4)}= \frac 23=\frac {2a}{3a} $.

The only stipulation is $a \ne -4$ and $x \ne -8$ and $y \ne -12$.

Answer 4

Fractions are classes of objects, so we first assume that $x/y$ has been reduced to the simplest representative of the class $p/q$. Then the question yields $$xy+8y=xy+12x,$$ which remains the same when $x=p$ and $y=q$, so that we have $$8q=12p.$$ From here we have that $p/q=2/3$, so that $x/y=pr/qr=2r/3r$ for any nonzero $r\in\mathbb{R}$, and we conclude that $x=2r$ and $y=3r$.