Absolute Value Problem College Algebra Level?

Question

Write Each Expression without absolute value symbols. I need the steps/explanation for the answers..Please keep your answers on the College Algebra Level. No Calculus (I don't need an embolism today lol).

1.) |x - 4| + |x + 5|, given 0
2.) |2x| - |x - 1|, given 0
The clearest and most detailed answer gets the Best Answer..Thanks

Answer 1

1.) Since x is between 0 and 1, x-4 would be somewhere between -3 and -4; therefore, |x-4| is between 3 and 4, or more specifically, |x-4| = 4-x. As for the other term, |x+5| = x+5. The whole expression would therefore be 4-x+x+5.

2.) Likewise, |x-1| = 1-x, and |2x| = 2x, so the whole expression is 2x - (1-x).

Final answers for both are left as an exercise for the reader. And remember: the only reason this works is because x is between 0 and 1.

Answer 2

1.)
First, general rule for absolute values is:
|a| = a or -a
ie., a > 0, or a < 0
Absolute value is a distance from zero, and a distance is never positive or negative. And positive or negative value tells us whether it's to the left or to the right.

In regard with this, when solving absolute value equations, with or without unknowns, you have to list all the possibilities given for the equation, then rule out those that are off limits (given with the restrictions for x, or the domain of x; read: x or whatever the unknown is).

Eg. |5| = 5 and -5, therefore you have to make two cases: one for |5| = 5, and other for |5| = -5

For your purpose:

|x - 4| + |x + 5|, given 0 Before we go any further, please note, that here x cannot have negative value (x>0), and cannot exceed 1; this will be very important when evaluating each case.
The restriction for x (0
Case 1:
|x-4|=x-4 > 0, |x+5|=x+5 > 0, 0 x>4, x>-5, 0 you can see that first case brings some contradictions: x cannot be between 0 and 1, and greater than 4, at the same time - that alone is enough to rule this case out
for your purpose: x>-5, which complies with the restriction, but due to the contradiction for x-4<0, we are stopping here

Case 2:
|x-4|=x-4 > 0, |x+5|=x+5 < 0, 0 x>4, x<-5, 0 This is an obvious rule-out.
As in the explanation for case 1, x cannot be >4 here, and this is enought to rule this case out. Another thing is that x cannot be less than -5, because it's minimum value is always greter than 0.

Case 3:
|x-4|=x-4 < 0, |x+5|=x+5 < 0, 0 x<4, x<-5, 0 If there were only |x-4|=x-4<0, it would be great, since x<4 (if x<1, it surely is <4)
here we also have a contradictory x<-5, which we already had in case 2, and this is enough to rule the case out.

Case 4:
|x-4|=x-4 < 0, |x+5|=x+5 > 0, 0 x<4, x>-5, 0 Finally, we have two viable values for our absolute values:
x can be <4, and x can be >-5.
Now, let's plug those in:
since |x-4|=x-4 < 0 works, we need to plug |x-4|=-(x-4)
since |x+5|=x+5 > 0, we need to plug |x+5|=x+5

|x-4|+|x+5|, 0 -(x-4)+(x+5) = -x + 4 + x + 5 (-x and x cancel each other)
= 9

Since all cases except 4 are ruled out,
|x - 4| + |x + 5| = 9, given 0
2.)
|2x| - |x - 1|, given 0 Again: since we have 2 absolute values, each of them 2 possible non-absolute values, therefore will examine 4 cases. Should we have 3 or more absolute values, we would have to examine as many combinations as such a case requires.

Again, x cannot have negative value here (x>0), and cannot exceed 1; this will be very important when evaluating each case.
Actually, this is the same situation as in 1.)

Solutions for 2.)
|2x| - |x - 1|, given 0
Case 1:
|2x|= 2x > 0, |x - 1|=x-1>0, given 0 x>0, x>1
x>0 complies with a given, and x>1 doesn't, which rules this case out

Case 2:
|2x|= 2x > 0, |x - 1|=x-1<0, given 0 x>0, x<1
this actually gives our "given", ie. 0 therefore we'll plug:
|2x|= 2x, |x - 1|=-(x-1)

|2x| - |x - 1|, given 0 2x-[-(x-1)] = 2x - (-x+1)
= 2x+x-1
= 3x-1

Case 3:
|2x|= 2x < 0, |x - 1|=x-1<0, given 0 x<0, x<1
x cannot be <0, therefore this alone rules this case out, although x<1 complies with a given

Case 4:
|2x|= 2x < 0, |x - 1|=x-1>0, given 0 x<0, x>1
neither x cannot be <0, nor x can be >1, which rules this case out

Only case 2 complies with a given 0 |2x| - |x - 1|=3x-1, given 0
And believe me, this is the way (you have to examine all the cases), 'cause this trick had cost me an A or B on my final exam in the high school math. One of the sources is listed below:

Answer 3

The first one is fairly simple. Whatever value is chosen, it will first reduce 4 in the first set of || by that amount. Next, it will increase the 5 in the next set of || by that same amount. In the end, it comes down to adding 5+4, which, obviously, equals 9. For example, if you choose .5 for X, you get

|-3.5| + |5.5| = 9

This works the same way for any number between 0 and 1.

The second is a bit more complicated. In essence, you first multiply the value of X by 2. This gives you 2X. Then, you basically take the value of X of of 1. Since X is always less than 1, when you then subtract the absolute value from 2x, you get 2X - (1-X). If you distribute the negative, this equals 3X - 1. For example:

|2x| - |x-1|

Now choose any value for X. I'll use .5, for ease of use.

|1| - |-.5| = .5

This is the same as 3(.5) - 1.

Answer 4

When you find something in absolute value signs, the result will always be positive, so the value inside the abs signs will either be positive or negative.
Quick example: |a| means that 'a' by itself might be positive or it might be negative (this is why it's difficult).

1)
|x - 4| is either equal to x - 4 (in the case that x > 4) or -(x - 4) (in the case that x < 4, go ahead and check, -(x - 4) will always be positive as long as 'x' is less than 4)
Well, it's already given that 0 < x < 1, so |x - 4| will always be the same as -(x - 4).
I hope it's easy to see that |x + 5| is always (x + 5).

So we have:
-(x - 4) + x + 5,
-x + 4 + x + 5,
4 + 5 = 9.

2)
In this problem, 'x' is always positive, so |2x| is always equal to 2x. The value of |x - 1| will either be (x - 1) or -(x - 1) depending on whether 'x' is greater or less than 1. Although 'x' is positive, it is always less than 1, so |x - 1| = -(x - 1) in any case.

Now we have:
2x - ( -(x - 1)),
2x - ( -x + 1),
2x + x - 1,
3x -1.

Feel free to email me for any clarifications

Answer 5

The trick here is that when you have a negative number you need to multiply by minus one to get the absolute value. When positive you can leave alone.

Q1)

In the range 0
(x+5) is positive so no change needed.

The equation is then (given the range)
4-x +x+5 = 9 , same as the book.

2)

Using similar arguments 2x is always positive so no change needed to get the absolute value.
x-1 is always negative so you need to multiply by minus one to make positive i.e |x-1| in this range is 1-x

This means the equation is 2x - (1-x) = 3x-1

Answer 6

1. In the range given, the first term is always negative, and the second always positive. Thus it is correct to rewrite the expression as: -(x - 4) + (x + 5) (why?) and the simplification is obvious.

2. Same idea, only now the second term is always negative. Hence, 2x - -(x - 1), or 2x + x -1, or 3x -1, as stated by the book.

Answer 7

1when x=0 !x-4!=!0-4!=4
!x+5!=!0+5!=5
so on adding !x-4!+!x+!=9
when x=1 !x-4!=!1-4!=3
!x+5!=!1+5!=6
on adding !x-4!+x+5!=9