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BillyZ
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D
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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
Untitled.jpg [ 61.01 KiB | Viewed 17331 times ]
A. 2
B. 3
C. 4
D. 5
E. 6
Attachment:
Untitled.jpg [ 61.01 KiB | Viewed 17331 times ]
13 Feb 2017, 12:27
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ziyuenlau
\(|3x-3|+|2x+8|<15\)
\(3*|x-1|+2*|x+4|<15\)
We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from -4" is less than 15.
Let's try to find the point where this distance is equal to 15.
........................ (-4) ...................................... (0) ........... (1) ..........................
The distance between -4 and 1 is 5. Thrice this distance is 15. So at the point x = -4, the sum will be 15. As we move to the right of -4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10.
What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now.
At x = 2, the sum becomes 3*1 + 2*6 = 15.
To the right of 2, the sum will keep increasing.
So the sum will be less than 15 between -4 and 2. This gives us 5 integer values (-3, -2, -1, 0, 1).
Answer (D)
13 Feb 2017, 08:16
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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.
\(2x+8 = 0\) 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.
