How Can You Solve This Modulus Equation with Square Roots and Absolute Values?

[chwala]

Homework Statement

$2|3x+4y-2|$+$3$$\sqrt{25-5x+2y}$=$0$

Relevant Equations

Modulus

I am trying to go through my old notes ...i came across this question,i do not have the solution.

Solve $2|3x+4y-2|$+$3$$\sqrt{25-5x+2y}$=$0$
Ok my approach on this;
$3$$\sqrt{25-5x+2y}$=$-2|3x+4y-2|$
$9(25-5x+2y)=4(3x+4y-2)^2$
$9(25-5x+2y)=4(9x^2+16y^2+24xy-12x-16y+4)$
$9(25-5x+2y)=4[(3x+4y)(3x+4y)-4(3x+4)+4]$
Let $p=(3x+4y)$ and $m=(2y-5x)$
Then it follows that
$9(25+m)=4(p^2-4p+4)$
$\frac {9}{4}$=$\frac{p^2-4p+4}{25+m}$
ok at this part i equated the numerators i.e
$(p-2)^2=9$
$→p-2=±3$, $p=5$ or $p=-1$
and on equating the denominator, i have $4=25+m$→$m=-21$

From this using $p=5$, and substituting in the problem, i end up with the simultaneous equation;
$3x+4y=5$
$-5x+2y=-21$,
$x≈3.615$ and $y≈-1.46$, i substituted these values into the original equation and they both satisfy the equation...
Any comments or better way of approaching the problem? Cheers guys

 

[ergospherical]

Solve $2|3x+4y-2|$+$3$$\sqrt{25-5x+2y}$=$0$

Both terms are non-negative so have to be zero independently, yes? The solution is the intersection of two straight lines.