Solving $4a+5b=27$ where $a,b$ are two different positive integers

Question

This problem is making my head spin:

The costs of equities of symbol A and symbol B (in dollars) are two different positive integers. If $4$ equities of symbol A and $5$ equities of symbol B together costs $27$ dollars, what is the total cost of $2$ equities of symbol A and $3$ equities of symbol B in dollars?

Select one from the follwing:

A. $15$

B. $24$

C. $35$

D. $42$

E. $55$

My attempt: I wrote down what's given first: $a>0$ and $b>0$. Then the two equations:

$4a + $5b = $27$ and $2a + 3b =$ ?.

I tried to solve it by elimination.

Say, $2a + 3b = 15$ and then solve for a and b given their assumption ($a>0$, $b>0$) holds.

But I'm going nowhere with this approach and would appreciate some help.

Answer 1

Let $2a+3b=x$. Then we have the system $4a+5b=27,2a+3b=x$. Then note that

$27=4a+5b=2(2a+3b)-b=2x-b\implies b=2x-27$.

So the solutions are $a=-\frac{5}{2}x+\frac{81}{2},b=2x-27$. Considering $a,b>0$, we have

$0<-\frac{5}{2}x+\frac{81}{2}\implies x<\frac{81}{5}=16.2$ and $0<2x-27 \implies x>27/2=13.5$. So $x$ is $14,15,$ or $16$. But if $x$ is even, $-\frac{5}{2}x$ is an integer and $a=-\frac{5}{2}x+\frac{81}{2}$ is therefore not. Therefore $x$ must equal $15$ if a solution exists. Plugging it in to our formulas for $a$ and $b$, $a=3,b=3$. But the condition says $a\neq b$, so no solution exists. If you ignore that condition (as the publisher seems to), then $15$ is the value of $2a+3b$.

Answer 2

4a + 5b = 27 can only have a = 3 and b = 3 as its solution, as we know that these are positive integers. There is no other possibility (to satisfy 'two different positive integers' requirement).

---

Do you have the source of this question? If from book, you may want to contact the publisher.