Diophantine equation

Question

How do I go about this question?

We sold respectively $m$ and $n$ units of items A and B. Assume $m\ge 1$, and the price of each unit of A is $\$3$ more than the price of each unit of B. Assume a total of $12$ units of both items was sold, and the price of the total was $1320$. Use a Diophantine equation to determine the price and the number of each the items A and B sold.

Do we form two equations of the form $$ m(x+3) +n(x)=1320 $$ and $$ m+n=12$$ and then go from there? It would be really appreciated if someone could solve this question as I don't know much about Diophantine equations and have been racking my brains with this one.

Answer 1

Expand the first equation and group the $x$ terms together so that you can substitute $m+n=12$: $$ x(m+n)+3m =1320\\ 12x +3m =1320 $$ For each choice of integer $m$, there is a corresponding price $x$ and $x+3$ for the two items such that the constraints are satisfied, but we are looking only for integer solutions for $x$. Continue simplifying. $$ x=\frac{1320-3m}{12}\\ x=110-\frac{m}{4} $$ This means for any integer $m$ in the range $1\leq m\leq 12$, if(and only if) $m$ is divisible by 4, you get an integer solution for $x$. So you can pick from $3$ different values of $m$.

Alternatively, you could substitute $m=12-n$ in the first equation, and proceed as shown above.

Answer 2

We sold respectively $m$ and $n$ units of items A and B. Assume $m\ge 1$,

$$ m \in \mathbb{N} \quad n \in \mathbb{N}_0 $$

and the price of each unit of A is $\$3$ more than the price of each unit of B.

$$ p_a = p_b + 3 $$

Assume a total of $12$ units of both items was sold,

$$ m + n = 12 $$

and the price of the total was $1320$.

$$ m p_a + n p_b = 1320 $$ We can use the prior information to formulate $$ (12 - n) (p_b + 3) + n p_b = 1320 \iff \\ 12 p_b - 3n = 1284 \quad (*) \\ $$ and have the condition $$ n \in \{ 0, \dotsc, 11 \} \quad (**) $$

Use a Diophantine equation to determine the price and the number of each the items A and B sold.

A linear Diophantine equation in two variables is of the form $$ a X + b Y = c $$ with integer $a, b, c, X, Y$. So I would assume that the prices $p_i$ are integers here and not dollars and cents or arbitrary real numbers.

Then our Diophantine equation is $(*)$ with the unknowns $p_b$ and $n$.

We can solve that equation by the usual algorithm and restrict the integer solutions to fit the condition $(**)$ and $p_b \ge 0$. From the feasible $p_b$ and $n$ we can infer the corresponding $m$ and $p_a$.

Solution:

The solution to $(*)$ is $$ (p_b, n) = (t + 107, 4t) \quad (t \in \mathbb{Z}) $$ considering $(**)$ we keep the solutions $t \in \{ 0, 1, 2 \}$, thus $$ (p_b, n) \in \{ (107, 0), (108, 4), (109, 8) \} $$ and $$ (m, n, p_a, p_b) \in \{ (12, 0, 110, 107), (8, 4, 111, 108), (4, 8, 112, 109) \} $$