Question

Find the least values of $x$ and $y$ which satisfy the equations:
$6x-13y=1$.

• A. $x=11,y=5$
• B. $x=9,y=11$
• C. $x=12,y=9$
• D. $x=4,y=6$

Answer 1

Answer: (A)

$x=11,y=5$

Given, $6x-13y=1$

$\Rightarrow x-\frac{13y}{6}=\frac{1}{6}$

$\Rightarrow x-2y-\frac{y}{6}=\frac{1}{6}$

$\Rightarrow x-2y-\frac{y+1}{6}=0$

We are solving for positive integers, so $x$ and $y$ are integers

$\Rightarrow \frac{y+1}{6}=$ integer

Let the integer be $p$

$\frac{y+1}{6}=p\Rightarrow y=6p-1$ .....(ii)

Substitute $y$ in (i), we get

$6x-13(6p-1)=1\Rightarrow 6x=78p-12\Rightarrow x=13p-2$ ......(iii)

From (ii) and (iii) we see that the values of $y$ and $x$ are negative for integer $p\le 0$, 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)

$\Rightarrow y=5,x=11$

So, the general solution is $x=13p-2,y=6p-1$ and the least value of $x$ and $y$ are $11$ and $5$ respectively.