I can’t figure out how to get the correct answer and hoping someone can pretty please show me how to solve:

-5│2+4x│= -32(x+3/4) -│x│+1

I can’t figure out how to get the correct answer and hoping someone can pretty please show me how to solve:

-5│2+4x│= -32(x+3/4) -│x│+1

Whew! That’s a bear. I tried it on paper, got it wrong. But I also think I’m approaching it the wrong way. If you’d like to try to work backwards from a solution, this might help (confirm that I got the equation right, though):

-5│2+4x│= -32(x+3/4) -│x│+1

Definition of |A|: |A| = A if A >= 0, and –A if A < 0

So |2+4x| = 2+4x if x >= -1/2, and –(2+4x) if x < -1/2

Similarly |x| = x if x >- 0, and –x if x < 0.

Case 1:

Assume x > 0:

Then both x and 2+4x are greater than 0, so |x|= x and |2+4x| = 2+4x.

Then -5|2+4x| = -32(x+3/4) – |x| + 1 becomes -5(2+4x) = -32(x+3/4) – x +1

Solving -5(2+4x) = -32(x+3/4) – x +1 gives x = -1.

But this contradicts our assumption that x > 0.

Therefore x must be less than 0.

Case 2:

By inspection x cannot be 0.

Case 3:

Assume that -1/2 <= x < 0.

Then |2+4x| = 2+4x, and |x| = -x.

Then -5|2+4x| = -32(x+3/4) – |x| + 1 becomes -5(2+4x) = -32(x+3/4) + x +1

Solving -5(2+4x) = -32(x+3/4) + x +1 gives x = -13/11 [unless I made an error somewhere].

But -13/11 < -1/2, contradicting our assumption that -1/2 <= x < 0.

Therefore x must be less than -1/2.

Case 4: Assume that x < -1/2.

Then |2+4x| = -(2+4x) = -2 – 4x, and |x| = -x.

Then -5|2+4x| = -32(x+3/4) – |x| + 1 becomes -5(-2- 4x) = -32(x+3/4) + x +1.

Solving -5(-2- 4x) = -32(x+3/4) + x +1 gives x = -11/17.

This does not contradict our assumption, and so is the correct answer.

I’m glad I saw this. This problem has me stumped too. Did I miss something or was this not taught in the class? I don’t remember doing anything like this in the homework, watching videos about it or reading about it.

It does seem an unusually difficult version of what’s presented later in Unit 1 – which shows how to solve equations with two absolute values but typically without other terms interfering, so you can set up two equations and see what happens.

Interested to hear comments from others on this problem.

It was definitely far beyond what was taught in class. Even with the answer worked out for me I had to really focus to follow it. But rjl was a real lifesaver for posting that!

I looked quickly through some recent exams and do see double absolute value questions, but typically only one example and not as complicated to solve as this one – more like what is in the course, where you just need to set up a couple equations and solve.

It’s worth going through the proof above until it makes sense – comfort with this will extend to comfort in other kinds of problems, too, particularly comfort with the logical underpinnings of the explanation. But in the context of an exam where time and anxiety can come into play, I’d consider the strategy of flagging the question and moving on, but then simplify as much (and as carefully) as you can and then start plugging in answer choices and evaluating.