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What is the solution of the equation \(|x - 2| + 2x = 4 - x\)?
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1 \)
E. \(2\)
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1 \)
E. \(2\)
25 Jun 2023, 04:29
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|x - 2| + 2x = 4 - x
or
|x - 2| + 3x = 4
Critical point of |x-2| is at x = 2
At x >= 2, |x-2|=x-2
At x < 2, |x-2| = 2-x
Therefore
x - 2 + 3x = 4 || or || 2 - x + 3x = 4
or
4x = 6 or x = 3/2 (but this does not match the original condition that x should be greater than or equal to 2)] or [ -2x = 2 or x = -1
or
|x - 2| + 3x = 4
Critical point of |x-2| is at x = 2
At x >= 2, |x-2|=x-2
At x < 2, |x-2| = 2-x
Therefore
x - 2 + 3x = 4 || or || 2 - x + 3x = 4
or
4x = 6 or x = 3/2 (but this does not match the original condition that x should be greater than or equal to 2)] or [ -2x = 2 or x = -1
02 Jul 2023, 09:16
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Fellow member TheBipedalHorse has explained how this question can be solved using equations.
Another very obvious approach is to enter value of x and then check. If done smartly, this will save you ton of time.
Always, start with the middle option i.e. C implying x = 0. However, this does not satisfy.
Now, shift to either B or D, let's say we go with D i.e. x = 1 and Bingo.
LHS = \(|x - 2| + 2x\) = 3
RHS = \(4- x\) = 3
Hence option D is the answer.
Another very obvious approach is to enter value of x and then check. If done smartly, this will save you ton of time.
Always, start with the middle option i.e. C implying x = 0. However, this does not satisfy.
Now, shift to either B or D, let's say we go with D i.e. x = 1 and Bingo.
LHS = \(|x - 2| + 2x\) = 3
RHS = \(4- x\) = 3
Hence option D is the answer.
Bunuel
What is the solution of the equation \(|x - 2| + 2x = 4 - x\)?
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1 \)
E. \(2\)
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1 \)
E. \(2\)
