- #1

chwala

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- 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

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

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