Solve this arithmetic question without algebra

Question

Dictionary A costs $64$ dollars, while dictionary B costs $54$ dollars. A teacher buys $59$ of the dictionaries altogether for 3426 dollars. How many of each did he/she buy?

64x + 54y = 3426 --- 1

x + y = 59 --- 2

Multiply 2 by 54,

54x + 54 y = 3186 --- 3

1 - 3,

10x = 240

x = 24

Sub. x into 2,

y = 35

I can solve this question with algebra but it is supposed to be solved with simple arithmetic only.

Answer 1

Dictionary A costs 64 dollars, while dictionary B costs 54 dollars.

This information provides you with information to calculate the cost: $$ 64 x + 54 y = c \quad (1) $$

A teacher buys $59$ of the dictionaries altogether.

This gives you the information relating $x$ and $y$. $$ x + y = 59 \quad (2) $$

Implicitly we have been given the constraints $$ x, y \in \mathbb{Z} \quad (3) $$ as we can buy or give only whole books and $$ x, y \ge 0 \quad (4) $$ as we buy only.

How many of each did he/she buy?

Let us have a look at the information in a graphical way:

The $x$-axis features the $x$-values, the $y$-axis the $y$-values.

The light blue areas are the half planes $x \ge 0$ and $y >0$. The red line $f$ represents all real values satisfying equation $(2)$.

Together with the constraints $(3)$ and $(4)$ all feasible solutions $(x, y)$ lie on the green line segment between $A=(0,59)$ and $B=(59,0)$. (We can not see the discrete gaps due to the scale of the image)

We see that this problem seems to allow many solutions: $$ X = \{ (0, 59), (1, 58), (2, 57), (3, 56), \dotsc, (58, 1), (59, 0) \} $$

As Paul Sinclair wrote in the comments, it is likely a piece of information is missing. The total cost $c$.

The image features the yellow line $g$, which stands for $c = 1442$ and equation $(1)$. The image suggests there is no intersection between $g$ and the line segment $AB$. So this $c$ was not a good choice.

Actually the task implicitly states that a sale happened.

So we now could check what $c$ values, which are likely to satisfy $c > 0$ and $c \in \mathbf{Z}$, intersect the set of feasible solutions on $AB$?

The other choices for $c$ each result in a line parallel to the yellow line $g$.

An intersection means a solution to $$ 64 x + 54 y = c \\ x + y = 59 $$ which gives: $$ y = 59 - x \\ c = 64 x + 54 (59 - x) = 10 x + 3186 $$

So all $60$ solutions are possible, each having a different total cost between $3186$ and $3776$.