Question

If $4x+3y=121$ find how many positive integer solutions are possible?

• A. $10$
• B. $1$
• C. $0$
• D. Cannot be determined

Answer 1

The correct option is A $10$

According to the given question we are required to find only non - negative integers.

We are given with the equation,

$\Rightarrow 4x+3y=121$

on rearranging we get something like,

$\Rightarrow 3y=121-4x$

$\Rightarrow 3y=120-6x+1+2x$

$\Rightarrow y=(\frac{120-6x}{3})+\left(\frac{1+2x}{3}\right)$

$\Rightarrow y=(40-2x)+\left(\frac{1+2x}{3}\right)$

we are required to find ordered pairs of $(x,y)$ that satisfy the above equation, also both should be positive.

Therefore, we get the following pairs.

$(1,39),(4,35),(7,31),(10,27),(13,23)$

$(16,18),(19,14),(22,11),(2.5,7),(28,3)$

$\Rightarrow$ There exist 10 positive integer solutions for the equation $4x+3y=121$

Hence, $\left(a\right)$ option is correct.