Algebraic Algebra

$3x + 4x = 7x$

$2y + 3y = 5y$

$7x + 5y = 11(3)= 33$

Now setting $y = 1$ , we find out which value of $x$ satisfies this equation:

$7(1) + 5(1) = 12 \neq 33$

$7(2) + 5(1) = 19 \neq 33$

$7(3) + 5(1) = 26 \neq 33$

$7(4) + 5(1) = 33$

Therefore, $x = 4$

Since we are asked to give our answer in the form $\overline{xy}$ , $\overline{xy} = \fbox{41}$

For $z=3$ , the equation reduces to $7x + 5y = 33$ . Taking modulo of $5$ on both sides, $2x \equiv 3 \text{ (mod 5)} \implies x \equiv 4$ . Putting $x=4$ in the equation, we have $28 + 5y = 33 \implies y = 1$ . Therefore, $\overline{xy} = \boxed {41}$ .

First we simplify the equation to get 7 $x$ + 5 $y$ = 11(3); 33

Then, we should think of $y$ as 1 to start

Input = $x$ ; Output = $x$ + 7 (Per Each Equation)

7(1) + 5(1) = 12

7(2) + 5(1) = 19

7(3) + 5(1) = 26

7(4) + 5(1) = 33

This shows that the answer is 41 since $x$ is 4 and $y$ is 1

$x=\dfrac {33-5y}{7}$

Only $y=1,x=4$ are the positive integer values that satisfy this equation. So the answer is $\boxed {41}$ .