guerrero25 wrote:
|x+3| – |4-x| = |8+x| How many solutions will this equation have?
A. 0
B. 1
C. 2
D. 3
E. 4
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) -8 <= x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) -3 <= x < 4 (x+3) - (4-x) = (8+x) --> x = 9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x >=4. (x+3) + (4-x) = (8+x) --> x = -1. We reject the solution because our condition is not satisfied (-1 is not more than 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.
I am sorry. I did not understand, how did you get 'when \(-3<x<4\), \(4-x\) is positive' and ' when \(x\geq{4}\), \(4-x\) is negative? Can you explain in details?