SOLVE :-

(3x +2y - 7)^2 +(5x + 7y - 13)^2 = 0, for real values of x and y

well, I'm not bad at maths, it's that this question is too challenging, so please HELP HELP HELP !!!!! (Worried)

- Mar 31st 2008, 08:26 AMice_syncerTough linear equations in 2 variable question, can you solve it :)
SOLVE :-

(3x +2y - 7)^2 +(5x + 7y - 13)^2 = 0, for real values of x and y

well, I'm not bad at maths, it's that this question is too challenging, so please HELP HELP HELP !!!!! (Worried) - Mar 31st 2008, 09:11 AMTheEmptySet
Since squares are always greater than or equal to zero, this equation will only have a solution when each of the squares are equal to zero.

So we need to solve the system.

$\displaystyle 3x+2y-7=0$

$\displaystyle 5x+7y-13=0$

Try either elimination or substition

I got

$\displaystyle x=\frac{23}{11} \mbox{ and }y=\frac{4}{11}$