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

• chwala
In summary, the conversation is about solving the equation ##2|3x+4y-2|##+##3####\sqrt{25-5x+2y}##=##0## by finding the intersection of two straight lines. The approach involves finding the values of x and y that satisfy the equation, and using simultaneous equations to solve for them. The solution is x≈3.615 and y≈-1.46.
chwala
Gold Member
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

Last edited:
chwala said:
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.

Delta2

## 1. What is a modulus equation?

A modulus equation is an equation that involves the absolute value or modulus function. This function returns the magnitude or distance of a number from zero, regardless of its sign.

## 2. How do you solve a modulus equation?

To solve a modulus equation, you need to consider two cases: when the number inside the absolute value is positive and when it is negative. For each case, you can set up an equation and solve for the variable. The resulting solutions will be the solutions to the original modulus equation.

## 3. Can a modulus equation have multiple solutions?

Yes, a modulus equation can have multiple solutions. This is because the absolute value function can return the same value for different inputs. For example, |x| = 5 has two solutions: x = 5 and x = -5.

## 4. Are there any special rules for solving modulus equations?

There are a few special rules for solving modulus equations. One is that if the absolute value is set equal to a negative number, the equation has no real solutions. Another rule is that if the absolute value is set equal to a positive number, the equation has two solutions, one positive and one negative.

## 5. Can a modulus equation have no solutions?

Yes, a modulus equation can have no solutions. This can happen when the absolute value is set equal to a negative number or when the equation has conflicting solutions for the two cases (positive and negative). In this case, the equation is considered to have no real solutions.

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