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\(3x-3 = 0\) then \(x = 1\)
\(2x+8 = 0\) then \(x = -4\)

there can be 3 range
\(x\leq{-4}\) | \(-4 < x < 1\) | \(x\geq{1}\)

1. When \(x\leq{-4}\), then both \(|3x-3|\) and \(|2x+8|\) will be negative.

\(-3x+3 -2x-8 < 15\)

\(-5x-5 < 15\)

\(x > -4\) (this is opposite to \(x\leq{-4}\). )

Hence \(x\leq{-4}\) is not a possibility.

2. When \(-4 < x < 1\), then \(|3x-3|\) would still be negative but \(|2x+8|\) will be positive. hence

\(-3x+3 +2x+8 < 15\)

-x+11 < 15

\(-x < 4\) i.e. \(x > -4\) (this lies within \(-4 < x < 1\)) so correct range.

3. When \(x\geq{1}\), then both \(|3x-3|\) and \(|2x+8|\) will be positive.

\(3x-3+2x+8 < 15\)

\(5x+5 < 15\)

\(x+1 < 3\)

\(x < 2\) this range is also fine because \(x\geq{1}\). so x must be 1.

So our range is \(-4 <\)\(x\geq{1}\), so howmany integers within this range?

-3, -2, -1, 0 and 1

Count is 5

Answer D.
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9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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For how many integer values of x, is \(|3x-3|+|2x+8|<15\)?

A. 2
B. 3
C. 4
D. 5
E. 6

\(positive:|3x-3|≥0…3x≥3…x≥1…negative:x<1\)
\(positive:|2x+8|≥0…2x≥-8…x≥-4…negative:x<-4\)
\(range:--(neg)--(-4)---(pos,neg)--(1)--(pos)---\)

\(x≥1:|3x-3|+|2x+8|<15…3x-3+2x+8<15…5x<10…x<2:1≤x<2=[1]\)
\(4≤x<1:…-3x+3+2x+8<15…-x<4…x>-4:-4<x<1=[-3,-2,-1,0]\)
\(x<-4:…-3x+3-2x-8<15…-5x<20…x>-4:invalid=x<-4\)

\(x=[-3,-2,-1,0,1]=5\)

Ans (D)
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-15 < 3x-3+2x+8 < 15
-15 < 5x + 5 < 15
-20 < 5x < 10
-4 < x < 2
X can be -3, -2, -1, 0 , 1, 2 . ( D )

­
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­The value hold true for a -3,-2,-1,0 and 1

All these can be solved very quickly. Total of 5, therefore answer is D
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One can always break the modulus and then proceed with the usual process as mentioned is other solutions but I find solving such problems by plotting the graph of the inequality. The visual is far better to interpret and avoids missing any values even if there are multiple such modulus.
Below is a descriptive solution for each step.

Posted from my mobile device
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Faster way of solving such questions would be to test values

Since the quesiton mentions that we need 'integer' values, the scope gets narrowed down

x = 0 --> |3|+|8 | < 15 yes
x = 1 --> |0|+|10| < 15 yes
x = 2 --> |3|+|12| < 15 no

So positive nos beyond 2 wont help

Lets try negetive numbers

x = -1 ---> |6| + |6| < 15 yes
x = -2 ---> |9| + |4| < 15 yes
x = -3 --->|12| + |2| < 15 yes
x = -4 --->|15| + |0| < 15 no

So numbers less than-4 wont help
Hence we have 5 numbers : Answer is D
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