Question

When one is subtracted from each of the two given numbers, their ratio becomes $2:3$ and if one is added to each , the ratio becomes $7:10$. find the numbers.

Answer 1

Step 1:Taking assumptions

Let $xandy$ be the two numbers

Step 2 : Applying given conditions and solving using cross multiplication method

According to the question,

Case 1 :

$$\begin{gathered} \frac{x-1}{y-1}=\frac{2}{3} \\ \Rightarrow 3(x-1)=2(y-1) \\ \Rightarrow 3x-3=2y-2 \\ \Rightarrow 3x-2y=3-2 \\ \Rightarrow 3x-2y=1.......................................\left(1\right) \end{gathered}$$

Case 2 :

$$\begin{gathered} \frac{x+1}{y+1}=\frac{7}{10} \\ \Rightarrow 10(x+1)=7(y+1) \\ \Rightarrow 10x+10=7y+7 \\ \Rightarrow 10x-7y=7-10 \\ \Rightarrow 10x-7y=-3.............................................\left(2\right) \end{gathered}$$

Step 3 : Making the coefficient of ‘x’ as same and solve using elimination method

$$\begin{gathered} equation\left(1\right)\times 10\Rightarrow 30x-20y=10.....................................\left(3\right) \\ equation\left(2\right)\times 3\Rightarrow 30x-21y=-9.....................................\left(4\right) \\ \left(3\right)-\left(4\right)\Rightarrow y=19 \end{gathered}$$

$$\begin{gathered} sub:y=19inequation1 \\ 1\Rightarrow 3x-2x19=1 \\ \Rightarrow 3x-38=1 \\ \Rightarrow 3x=39 \\ \Rightarrow x=\frac{39}{3}=13 \end{gathered}$$

$\therefore$ the required numbers are x = 13 & y=19

* Verification

$$\begin{gathered} \frac{x-1}{y-1}=\frac{13-1}{19-1}=\frac{12}{18}=\frac{2}{3} \\ \frac{x+1}{y+1}=\frac{13+1}{19+1}=\frac{14}{20}=\frac{7}{10} \end{gathered}$$