Question

what are the possible values of x
x^2 + 5xy + 4y^2 = 55
x^2 + 7xy + 12y^2 = 99​

Answer 1

1st equation $\sf{(x+y)(x+4y)=55}$

2nd equation $\sf{(x+3y)(x+4y)=99}$

Two equations are both quadratic.

Apart from the question, let's think about how we solve linear equations in two variables.

The idea is to subtract like terms then solve for one variable.

Similarly, we can lower the degree of equations and use this idea.

Divide 1st by 2nd

$\implies\sf{\dfrac{x+y}{x+3y} =\dfrac{5}{9} }$

$\implies\sf{9(x+y)=5(x+3y)}$

$\implies\sf{9x+9y=5x+15y}$

$\implies\sf{4x=6y}$

So, $\sf{2x=3y}$ is a condition between two variables.

Multiply the whole equation by 4 and distribute factors

1st equation $\sf{(2x+2y)(2x+8y)=220}$

Given condition makes $\sf{(3y+2y)(3y+8y)=220}$

$\implies\sf{5y\times 11y=220}$

$\implies\sf{55y^2=55\times4}$

$\implies\sf{y^2=4}$

$\therefore\sf{y=\pm2}$

The solution becomes $\sf{x=\pm3}$ because of the given condition.