carcass wrote:
Here you can:
a) test each answer choice OR
b) take the algebra way
Both are fast but it depends how you are comfortable but if you start thinking about the positive side of absolute value ---> \(x - 3 + x + 1 + x = 10\) ---> \(3x = 12 x = 4\) OR\(- x + 3 -x - 1 - x = 10\) ----> \(- 3x = 8\) NOT POSSIBLE
Only D takes in account. in no more than 50 seconds - 60 at maximum
regards
Going the algebra way, you have missed two cases. There are 3 transition points: -1, 0, 3
When x > 3,
\(|x - 3| + |x + 1| + |x| = 10\)
\(x - 3 + x + 1 + x = 10\)
x = 4
Satisfies
When 0 < x < 3
\(|x - 3| + |x + 1| + |x| = 10\)
\(-(x - 3) + (x + 1) + x = 10\)
x = 6 (Not possible since 0 < x < 3)
When -1 < x < 0
\(|x - 3| + |x + 1| + |x| = 10\)
\(-(x - 3) + -(x + 1) + x = 10\)
x = -8 (Not possible since -1 < x < 0)
When x < -1
\(|x - 3| + |x + 1| + |x| = 10\)
-(x - 3) - (x+1) - x = 10
x = -8/3
Satisfies
Better is the number line method I discussed in this post:
http://www.veritasprep.com/blog/2011/01 ... s-part-ii/The question has been picked from here. Mind you, I have discussed two different questions:
1. For what value of x, is |x – 3| + |x + 1| + |x| = 10?
2. For how many values of x, is |x – 3| + |x + 1| + |x| = 10?
The method to be used is different in the two cases. As Ron mentioned above, just plug in the options in the first question. In the second question, you can use the number line to quickly solve it. Check out the post to avoid confusion.
Karishma - could you elaborate on the transition points please? What are they and why it is helpful to find them. Thanks in advance.