Bilinear integer equation

[juantheron]

If $x,y$ are integer ordered pair of $2x^2-3xy-2y^2 = 7,$ Then $\left|x+y \right| = $

My Try:: Given $2x^2-3xy-2y^2 = 7\Rightarrow 2x^2-4xy+xy-2y^2 = 7$

So $(2x-y)\cdot (x-2y) = 7.$

Now How can I solve after that

Help me

Thanks

 

[Aryth1]

If $x,y$ are integer ordered pair of $2x^2-3xy-2y^2 = 7,$ Then $\left|x+y \right| = $

My Try:: Given $2x^2-3xy-2y^2 = 7\Rightarrow 2x^2-4xy+xy-2y^2 = 7$

So $(2x-y)\cdot (x-2y) = 7.$

Now How can I solve after that

Help me

Thanks

You are just a little bit off. You can check your work. Let's try that here.

$(2x-y)\cdot (x-2y) = 2x^2 - 4xy -xy + 2y^2 = 2x^2 -5xy + 2y^2 \neq 2x^2 - 3xy - 2y^2$.

Knowing this, what do you think the correct factorization is?

 

[Opalg]

Once you have got the correct factorisation sorted out, notice that if the product of two integers is $7$, then one of them must be $\pm1$ and the other one must be $\pm7$.

 

[juantheron]

Thanks Aryth, opalg Got it.

 

[CellCoree]

I would suggest using algebraic techniques to solve for the ordered pair $(x,y)$ in the equation $2x^2-3xy-2y^2 = 7$. One approach could be to factor the left-hand side of the equation as you have already done, and then consider all possible combinations of factors of 7 to find potential solutions. Another approach could be to use the quadratic formula to solve for $x$ in terms of $y$ and then substitute that expression into the equation to solve for $y$. Once you have found the values of $x$ and $y$, you can plug them into the expression $|x+y|$ to find the final answer. It is important to also check your solutions to make sure they satisfy the original equation.