Question

Find the general solution in positive integers, and the least values of $x$ and $y$ which satisfy the equations:
$5x-7y=3$.

• A. $x=-1,y=1$
• B. $x=2,y=1$
• C. $x=1,y=2$
• D. $x=2,y=2$

Answer 1

Answer: (B)

$x=2,y=1$

$5x-7y=3$ ......(i)

$\Rightarrow x-\frac{7y}{5}=\frac{3}{5}\Rightarrow x-y-\frac{2y}{5}-\frac{3}{5}=0\Rightarrow x-y-\frac{2y+3}{5}=0$

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

$\frac{2y+3}{5}=$ integer

Multiplying by $3$, we get

$\Rightarrow \frac{6y+9}{5}=$ integer

$\Rightarrow y+1+\frac{y+4}{5}=$ integer

Let the integer be $p$

$\frac{y+4}{5}=p\Rightarrow y=5p-4$ ......(ii)

Substituting in (i), we eget

$5x-7(5p-4)=3\Rightarrow 5x=35p-25\Rightarrow x=7p-5$ .....(ii)

We see from (ii) that $y<0$ for integer $p<1$

So, the minimum value of $p$ is $1$

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

$\Rightarrow y=1,x=2$

So, the least value of $x$ is $2$ and $y$ is $1$.