# Thread: Solving a system involving products

1. ## Solving a system involving products

The following is from a list of problems that, as far as I know, all have solutions. Unfortunately, I don't have a list of solutions.

Solve the following system, where $\displaystyle x > y$:

$\displaystyle x + y + xy = 19$

$\displaystyle xy(x + y) = 84$

The answer is $\displaystyle x=4, y=3$. I don't know if there is another solution, and I got it simply by inspection - is there an algebraic method to get the answer (or a reason why it isn't algebraically solvable)?

Thanks!

2. Hello, Mathnasium!

I found an approach.
Don't know if it's any better than Brute Force algebra.

Solve the following system, where $\displaystyle x > y$:

$\displaystyle \begin{array}{ccc}x + y + xy &=& 19 \\ xy(x + y) &=& 84\end{array}$
Let $\displaystyle a\:=\:x+y,\;\;b \:=\:xy$

Then we have: .$\displaystyle \begin{array}{cccc}a + b &=& 19 &{\color{blue}[1]} \\ ab &=& 84 &{\color{blue}[2]} \end{array}$

From [1], we have: .$\displaystyle b \:=\:19-a$

Substitute into [2]: .$\displaystyle a(19-a)\:=\:84\quad\Rightarrow\quad a^2-19a+84\:=\:0$

Hence: .$\displaystyle (a-7)(a-12)\:=\:0\quad\Rightarrow\quad a \:=\:7,\,12 \quad\Rightarrow\quad b \:=\:12,\,7$

If $\displaystyle a=7,\:b = 12$, we have: .$\displaystyle \begin{array}{ccc}x+y &=&7\\ xy &=&12 \end{array}$

. . which has the solution: .$\displaystyle \boxed{x = 4,\:y = 3}$

If $\displaystyle a= 12,\:b=7$, we have: .$\displaystyle \begin{array}{ccc}x+y &=&12 \\ xy&=& 7\end{array}$

. .which has the solution: .$\displaystyle \boxed{x \:=\:6+\sqrt{29},\;\;y \:=\:6-\sqrt{29}}$