How Do You Tackle This Modulus Equation Challenge?

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To solve the modulus equation |12x - 9| = |3 - 4x| + 2x - 1, it's essential to analyze the conditions based on the critical point x = 3/4. For x ≥ 3/4, |12x - 9| is positive, while |3 - 4x| is negative, leading to the equation 12x - 9 = 4x - 3 + 2x + 1. For x < 3/4, the roles reverse, requiring the consideration of both cases for the absolute values. The solution for x > 3/4 yields x = 7/6, which is confirmed as correct. Understanding these cases is crucial for accurately solving modulus equations.
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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. :/
 
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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:

12x-9\geq 0

x\geq 3/4

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

For the other inequality, at x=3/4 we have 3-4x=0 and for any x\geq 3/4 we have 3-4x\leq 0 So this means for all x\geq 3/4 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 x&lt;3/4.
 
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?
 
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 a\geq 0 then you leave that as a, when a&lt;0 you change that to -a (since the negative of a negative number is positive). But for the other inequality, when a+1\geq 0 or in other words, a\geq -1 then that absolute value is positive and thus left as a+1, so when a&lt;-1 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 -1&lt; a&lt; 0? 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.
 
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?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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