|x+3| - |4-x| = |8+x|. How many solutions does the equation

|x+3| – |4-x| = |8+x| How many solutions will this equation have?

A. 0
B. 1
C. 2
D. 3
E. 4

I solved it very differently and got the right answer. Not sure whether it was a fluke.

I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.

Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.

Is this approach correct?

Could you explain the same by solving with just one of the examples but with |4-x|

I understood the concept you have explained... but I am confused how would it be solved if I dont convert it to |x-4|..

Thanks!

Responding to pm.

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

STEP BY STEP SOLUTION:

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<-8\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is negative. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=-(8+x)\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =-(8+x)\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x<-8\)).

2. When \(-8\leq{x}\leq{-3}\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =8+x\): --> \(x=-15\). This solution is NOT OK, since \(x=-15\) is NOT in the range we consider (\(-8\leq{x}\leq{-3}\)).

3. When \(-3<x<4\), then \(x+3\) is positive, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (4-x) =8+x\): --> \(x=9\). This solution is NOT OK, since \(x=9\) is NOT in the range we consider (\(-3<x<4\)).

4. When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).

Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).

Answer: A.

Hope it's clear.

Hi Bunuel,

I am getting confused at the sign changes. Please explain a bit clearly.

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Now we have four ranges:

As per property...................... |x| < 0 ... | some expression | = - some expression ---> Property 1
|x| > 0 ... | some expression | = some expression ----> Property 2.

Confirm the below:
Now case 1 : x < -8 then |x+3| = - (x+3) and |x-4| = - ( x-4) ( this has to -ve but you mentioned as positive ) and |x+8| = - (x +8 )

and confused for the other three ranges as well on how the expressions got +ve and -ve signs.
What all need to be checked to give positive and negative signs to the expressions, as you mentioned in the solution for the above four ranges.

|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.

Will you please explain why you didn't take -(x+3) + (4-x)

Why you counted -(4-x)

You do it in the proper way. This method is called the "critical values" method. And once you have the critical values (by doing each absolute term equal to zero), you have to place them in the Real numbers line to make all the possible intervals. Then you just do the intervals as follows: x<lowest number in your real line, and then you take the intervals from each critical value to before the next one: i.e. x<-8, -8<=x<-3, -3<=x<4, x<=4. Therefore, you are getting all the possible intervals in the real line, and splitting the intervals from one critical value (including it) to before the next critical value (not including it).

Then, as you have done, you just set the predominant sign for each term under each condition, you solve the equation, and finally you check if the result saqtisfies the condition.

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I find the above technique easier for these problems.

Can someone please confirm if my below approach is correct :

The roots/criticial values for the given equations are : -8,-3,4.

Now placing these values on the no. line we have four ranges & checking for each ranges :

1. For x greater than 4

Randomly selecting any value greater than 4 , lets say x = 10

|x+3| - |4-x| = |8+x|

|10+3| - |4-10| is not equal to |8+10|

So any value greater than 4 doesnot satisfy the equation.

2. For x between -3 & 4

Let x = 2
|x+3| - |4-x| = |8+x|

|2+3| - |4-2| is not equal to |8+2|

So any value between -3 & 4 doesnot satisfy the equation.

3. For x between -8 & -3,

Let x = -2

|x+3| - |4-x| = |8+x|

|-2+3| - |4-(-2)| is not equal to |8+(-2)|

So any value between -8 & -3 doesnot satisfy the equation.

4. For x lesser than -8,

Let x = -9

|x+3| - |4-x| = |8+x|

|-9+3| - |4-(-9)| is not equal to |8+(-9)|

So any value less than -8 doesnot satisfy the equation.

So the total solution is 0.

Please confirm if my above approach is correct .

Thanks
Kshitij

|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The only point is that we don't include the transition points twice.

Hope the role of '=' sign is clear.

hi mam

in "a" when assumed that x is less than -8, the RHS modulus (8+x) is correctly negated. I am okay with this operation. Side by side, however, the LHS modulus (x + 3) is also negated...

what is the main logic underlying this negation ..? is this such that to make the LHS negative, some number should be subtracted from some bigger negatives..? if so, how can we conclude that (x+3) is bigger in value than is (4-x) ..?

thanks in advance ...

Hello Bunuel, Can you explain me the following.

When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?

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Hello Bunuel, Thanks for your reply.
I can understand the solution for x>4 , which turns out to be x=-1 and not in the range we considered.
But I am not able to understand the part when we take boundary value say x=4 , then 4-x=4-4=0 , and we know |x|=x ,x>=0, so why are we not considering this point in solution. So why are we considering |4-x| =-(4-x) only for x>4 ?
I think I am missing something here .Please explain.

Hey Everyone,

I had a query.

|x+3| – |4-x| = |8+x|==> -|4-x|=|x+8|-|x+3|
Now the RHS of the equation is always 5 for all values of x
therefore,

-|4-x|=5

Solving for x we get x=-1 or x= 9

if x= -1
Equation==> |-1+3|-|4-(-1)|=|8-1|==> |2|-|5|=|7| LHS=/=RHS so not applicable

if x=9
Equation==> |9+3|-|4-9|=|8+9|==> |12|-|-5|=|17| LHS=/=RHS so not

There fore Option A is correct.

What is the flaw in this solution. and if there is no flaw then can I replicate this for other such questions? In what situations would this logic not work and why? Can anyone help?

Many thanks in advance