Question

Solve in positive integers:
$7x+12y=152$.

• A. $x=20,y=1$
• B. $x=12,y=8$
• C. $x=8,y=8$
• D. $x=16,y=7$

Answer 1

Answer: (C)

$x=8,y=8$

$7x+12y=152$ .....(i)

$\Rightarrow x+\frac{12y}{7}=\frac{152}{7}\Rightarrow x+y+\frac{5y}{7}=21+\frac{5}{7}\Rightarrow x+y+\frac{5y-5}{7}=21$

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

$\Rightarrow \frac{5y-5}{7}=$ integer

Multiplying by $3$, we get

$\frac{15y-15}{7}=$ integer

$\Rightarrow 2y-2+\frac{y-1}{7}=$ integer

$\Rightarrow \frac{y-1}{7}=$ integer

Let the integer be $p$

$\frac{y-1}{7}=p\Rightarrow y=7p+1$ .....(ii)

Substituting $y$ in (i)

$7x+12(7p+1)=152\Rightarrow 7x=84p-140\Rightarrow x=20-12p$

From (ii) we see that $y<0$ for $p<0$ and from (iii) we see that $x<0$ for $p>1$.

So the values of $p$ can be $0,1$

Substituing $p$ in (ii)

$\Rightarrow y=1,8$

Substituing $p$ in (iii)

$\Rightarrow x=20,8$

So the complete solution set of positive integers is

$\{\begin{array}{c}x=20,8\\ y=1,8\end{array}$