Question

Find the least values of $x$ and $y$ which satisfy the equations:
$8x-21y=33$.

• A. $x=10,y=6$
• B. $x=12,y=3$
• C. $x=14,y=5$
• D. $x=9,4$

Answer 1

The correct option is B $x=12,y=3$

Let $8x-21y=33$ ....(i)

$\Rightarrow x-\frac{21y}{8}=\frac{33}{8}\Rightarrow x-2y-\frac{5y}{8}=4+\frac{1}{8}\Rightarrow x-2y-\frac{5y+1}{8}=4$

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

$\Rightarrow \frac{5y+1}{8}=$integer

Multiplying by $5$, we get

$\Rightarrow \frac{25y+5}{8}=$ integer

$\Rightarrow 3y+\frac{y+5}{8}=$ integer

$\Rightarrow \frac{y+5}{8}=$ integer

Let the integer be $p$

$\frac{y+5}{8}=p\Rightarrow y=8p-5$ ......(ii)

Substituting $y$ in (i), we get

$8x-21(8p-5)=338x=168p-72x=21p-9$ ......(iii)

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

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

Substituting $p=1$ in (ii) and (iii), we get

$\Rightarrow y=3,x=12$

So, the general solution is $x=21p-9,y=8p-5$ and least value of $x$ and $y$ are $12$ and $3$ respectively.