Question

Find the least values of $x$ and $y$ which satisfy the equations:
$77y-30x=295$.

• A. $x=32,y=27$
• B. $x=21,y=34$
• C. $x=47,y=35$
• D. $x=46,y=32$

Answer 1

The correct option is C $x=47,y=35$

Given equation is $7y-30x=295$

$\Rightarrow \frac{77y}{30}-x=\frac{295}{30}$

$\Rightarrow \frac{17y}{30}+2y-x=9+\frac{25}{30}$

$\Rightarrow \frac{17y-25}{30}+2y-x=9$

As $x$ and $y$ are positive integers,

$\Rightarrow \frac{17y-25}{30}$integer

Multiplying by $23$, we get

$\Rightarrow \frac{391y-575}{30}=$integer

$\Rightarrow 13y-19+\frac{y-5}{30}=$integer

$\Rightarrow \frac{y-5}{30}=$integer

Let the integer be $p$

$\frac{y-5}{30}=p$

$\Rightarrow y=30p+5$ ......(ii)

Substituting $y$ in (i)

$\Rightarrow 77(30p+5)-30x=295$

$\Rightarrow 2310p+385-30x=295$

$\Rightarrow 30x=2210p-90$

$\Rightarrow x=77p-30$ ........(iii)

We can see from (ii) and (iii) that value of $y$ and $x$ is negative for integer $p<1$, which is not possible as we are solving for positive integers.

So, the least value of $p$ is $1$.

Substituting $p=1$ in (ii) and (iii)

$\Rightarrow y=35,x=47$

So, the general solution is $c=77p-30,y=30p+5$ and the least value of $x$ and $y$ are $47$ and $35$ respectively