How Do You Tackle This Modulus Equation Challenge?

In summary, when solving the equation |12x -9| = |3 -4x| +2x -1, we need to consider two cases: when x is greater than or equal to 3/4 and when x is less than 3/4. When x is greater than or equal to 3/4, we solve the equation 12x -9 = 4x -3 +2x +1. When x is less than 3/4, we solve the equation -12x +9 = 3 -4x +2x -1. The final solution will be the combination of these two cases.
  • #1
Azure Skye
3
0
I have to solve the equation: |12x -9| = |3 -4x| +2x -1

I was wondering if it's supposed to be like:

12x -9 = 3 -4x +2x -1 and -12x +9 = 3 -4x +2x -1 ?

Or am I supposed to consider the positive and negative of 3 -4x also?
I didn't attend lessons for this and I know little about modulus functions. :/
 
Physics news on Phys.org
  • #2
Actually what you do is you find what values of x make one of the absolute values more than zero, so let's do that now:

[tex]12x-9\geq 0[/tex]

[tex]x\geq 3/4[/tex]

Therefore for all [tex]x\geq 3/4[/tex] that inequality is positive.

For the other inequality, at x=3/4 we have 3-4x=0 and for any [tex]x\geq 3/4[/tex] we have [tex]3-4x\leq 0[/tex] So this means for all [tex]x\geq 3/4[/tex] we have the first inequality is more than zero, and the second is less than zero (so we take the negative of it). And things are just reversed when we consider when [tex]x<3/4[/tex].
 
  • #3
So does the inequality formed from |12x -9| applies to |3 -4x|?

Then do I end up with something like 12x -9 = 0 +2x -1?
 
  • #4
Yes it does apply to the other inequality.

If you have |a|=|a+1| then you need to look at one of the absolute values. For the first, when [itex]a\geq 0[/itex] then you leave that as a, when [itex]a<0[/itex] you change that to -a (since the negative of a negative number is positive). But for the other inequality, when [itex]a+1\geq 0[/itex] or in other words, [itex]a\geq -1[/itex] then that absolute value is positive and thus left as a+1, so when [itex]a<-1[/itex] it becomes -(a+1)=-a-1.

So this means when a<-1 we have both absolute values are negative so we solve -(a)=-(a+1) and when a>0 we have both are positive, so we solve a=a+1. What about for [itex]-1< a< 0[/itex]? Well we have that the first is negative, and the second is positive so we solve -a=a+1

The same idea applies to your problem, so no, you don't solve 12x-9=0+2x-1. Use the idea of cases when x>3/4 and x<3/4 (this problem is easier since you only need two cases). Also, you can add in the value of x=3/4 later or just include it into each inequality, it doesn't really matter.
 
  • #5
Does this means that because |12x -9| is x > 3/4 and |3 -4x| is x< 3/4, when 12x -9 is positive, 3 -4x is negative and vice versa?

So when I consider x > 3/4, I get
12x -9 = 4x -3 +2x +1
x = 7/6

Is that correct or wrong?
 

FAQ: How Do You Tackle This Modulus Equation Challenge?

What is a modulus function equation?

A modulus function equation is an equation that contains an absolute value expression. It is used to calculate the distance of a number from zero on the number line. It is represented by the symbol |x| and is always positive.

How do you solve a modulus function equation?

To solve a modulus function equation, you need to consider two cases: when the value inside the absolute value is positive and when it is negative. If the value is positive, you can simply solve the equation as normal. If the value is negative, you need to make it positive by multiplying it by -1. Then, solve the equation as normal. The resulting solutions will be the distance from zero on the number line.

What are some real-life applications of modulus function equations?

Modulus function equations are used in many fields, including physics, engineering, and economics. They are used to calculate distances, velocities, and accelerations in motion problems. They are also used in calculating the magnitude of electric and magnetic fields in physics, and for determining the optimal solutions in game theory in economics.

Are there any restrictions when working with modulus function equations?

Yes, there are some restrictions when working with modulus function equations. The value inside the absolute value cannot be negative, as this would result in an imaginary solution. Additionally, when solving an equation with multiple modulus functions, the resulting solutions must be within the range of all the modulus functions involved.

Can modulus function equations be graphed?

Yes, modulus function equations can be graphed. The graph of a modulus function is typically a V-shaped graph, with the vertex at the origin. The slope of the graph will be positive on one side and negative on the other side. The graph can also be shifted vertically and horizontally using transformations.

Similar threads

Back
Top