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If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $\displaystyle \frac{5}{8}$ and if the numerator and the denominator of the same fraction are each increased by $1$, the fraction becomes equal to $\displaystyle \frac{1}{2}$. Find the fraction.

I tried, Let the numerator of the fraction be $x$, Let the denominator of the fraction be $y$ Therefore, Original fraction is $\displaystyle \frac{x}{y}$.

$$ \left\{ \begin{eqnarray*} \frac{x+2}{y+1}=\frac{5}{8} \\[2mm] \frac{x+1}{y+1}=\frac{1}{2} \\ \end{eqnarray*} \right. $$

What should I do now?

Roby5
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Abhishekstudent
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3 Answers3

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That's a fine start. Now you've got two equations in two unknowns. $$\frac{x+2}{y+1} = \frac 58\tag{1}$$

$$\frac{x+1}{y+1} = \frac 12 \tag{2}$$

Now, we have to assume that $y+1 \neq 0$, given the information that's impossible, so we can take $(1)$ for example, and "cross multiply" to get: $$ 8(x+2) = 5(y+1) \iff \color{blue}{y + 1} = \frac{8(x+2)}{5}\tag{1}$$

From the second equation, we have

$$2(x+1) = \color{blue}{y+1}\tag{2}$$

That means $\color{blue}{y+1 = y+1}$, so that $$\frac {8(x+2)}5 = 2(x +1)\tag{3}$$

Now solve for $x$: $$\frac{8(x + 2)}5 = 2(x+1)\iff 5(x+1) = 4(x+2) \iff 5x+5 = 4x + 8 \iff x = 3$$ Then use $x=3$ to solve for $y$, using the fact that $y + 1 = 2(x + 1) \iff y = 2x + 1\implies y = 2(3) + 1 = 7$.

So the original fraction $$\frac xy = \frac 37$$

amWhy
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    Also note that the second equation is in some sense "unnecessary" to solve the problem. 5-2 is 3, and 8-1 is 7, so 3/7 is easily found as the "most obvious" fraction that satisfies the first requirement. It also satisfies the second, so we're done without needing to use simultaneous equations. Of course, that isn't a general way to solve problems of this kind :-) – Steve Jessop Jul 28 '14 at 17:38
  • abhishekstudent - Is this clear now? – amWhy Jul 28 '14 at 19:03
2

$\dfrac{x+2}{y+1}=\dfrac{5}{8}$ and $\dfrac{x+1}{y+1}=\dfrac{1}{2}=\dfrac{4}{8}$

Dividing first equation by second equation you would get

$$\dfrac{x+2}{x+1}=\dfrac{5}{4}=\dfrac{3+2}{3+1}$$

This should be easy now..

1

We have the equations $$\frac{x+2}{y+1}=\frac{5}{8}, \ \frac{x+1}{y+1}=\frac{1}{2}=\frac{4}{8}$$ Hence $$x+1=4, \ y+1=8 \ \implies x=3, \ y=7$$ We are allowed to do so since $(x+2)-(x+1)=5-4$

SuperAbound
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