# Modulus equation

Gold Member
Homework Statement:
solve the modulus equation ##2|3x+4y-2|+3(25-5x+2y)^{1/2} = 0##
Relevant Equations:
modulus equations
##3√(25-5x+2y) = -2|3x+4y-2|##
→##9(25-5x+2y)=4[(3x+4y)^2-4(3x+4y)+4]##
→##9 (25+m)= 4[p^2-4p+4]## where ##p= 3x+4y## & ##m= 2y-5x##
##\frac{(p-2)^2}{25+m}##=## \frac{9}{4}##

Last edited:

Homework Helper
Hello again,

You have one equation with two unknowns, and you work it around to one equation in two unknowns. That does not help very much.

Are you aware that the term 'solving' is not applicable if there are more unknowns than equations ? E.g. ##x+y=4## is not solvable. All one can do is work it around to e.g. ##y = 4-x##.

Is that the aim of this exercise ?

Then: you square left and right sides of an equation. Do you realize that introduces unwanted solutions ? Look at e.g. the equation ##z = -2##, a single solution. Whereas ##z^2 = 4## has two solutions ?

Having said all this, there is something remarkable with your first equation:
$$2|3x+4y-2|+3(25-5x+2y)^{1/2} = 0$$
(you want curly brackets around the exponent 1/2 )

You have two terms that are non-negative and should add up to zero.
What is the only way in which that can be so ?

ehild
Mentor
(you want curly brackets around the exponent 1/2 )
Fixed in original post.

Gold Member
a colleague was able to solve it for me...by coming up with simultaneous equations:

##|6x+8y-4| =√(225-45x+18y)##
##|6x+8y-4| =-√(225-45x+18y)##...on solving the simultaneous by subtracting both sides of the equation,

##2√(225-45x+18y)=0##
##45x-18y=225## also using
##-6x-8y+4= 6x+8y-4## we get
##3x+4y=2##
now solving,
##45x-18y=225##
##3x+4y=2##
##x=4## ##y=-2.5##
is there another approach? or this is the only way? just thinking of the approach that i had used...

Definitely. $$a+b = 0 \quad \&\quad a \ge 0 \quad \& \quad b\ge 0 \quad {\bf \Rightarrow} \quad a = 0 \quad \& \quad b = 0$$
and now you have two equations with two unknowns:$$25-5x+2y = 0 \quad \& \quad 3x+4y-2 = 0$$with, of course, the same answer -- without multiplying within the | | and underneath the square root