Question

Find the least values of $x$ and $y$ which satisfy the equations:
$19y-23x=7$.

• A. $2,3$
• B. $3,4$
• C. $4,5$
• D. $2,1$

Answer 1

Answer: (B)

$3,4$

Let $19y-23x=7$ ......(i)

$\Rightarrow y-\frac{23x}{19}=\frac{7}{19}\Rightarrow y-x-\frac{4x}{19}=\frac{7}{19}\Rightarrow y-x-\frac{4x+7}{19}=0$

As $x$ any $y$ are positive integers

$\Rightarrow \frac{4x+7}{19}=$integer

Multiplying by $5$, we get

$\Rightarrow \frac{20x+35}{19}=$ integer

$\Rightarrow x+1+\frac{x+16}{19}=$ integer

$\Rightarrow \frac{x+16}{19}=$ integer

Let the integer be $p$

$\frac{x+16}{19}=p\Rightarrow x=19p-16$ ........(ii)

Substituting $x$ in (i), we get

$19y-23(19p-16)=7\Rightarrow 19y=437p-361\Rightarrow y=23p-19$ ......(iii)

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

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

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

$\Rightarrow x=3,y=4$

So, the general solution is $x=19p-16,y=23p-19$ nad the least value of $x$ and $y$ are $3$ and $4$ respectively.