Solve for x: |x-3| |2x-1| A) x∈ (-∞, 4/3) ∪ (2, ∞)

Question

Solve for x: |x-3|> |2x-1|

A) x∈ (-∞,  \frac{4}{3} ) ∪ (2, ∞)
B) x∈ (-2,  \frac{4}{3} )
C) x∈ (- \frac{4}{3} , 2)
D) x∈ ( \frac{4}{3} , 2)
E) x∈ (-∞, -2) ∪ ( \frac{4}{3} , ∞)

Himanshu453 · Answer

Divide the number line into 3 segments
x<1/2, 1/2<X<3, X>3
as at these points the value of the function becomes equal to zero.

x<1/2
both the sides of the equation are negative, thus
-(x-3)>-(2x-1)
.......x>-2

1/2<X<3
fot this part the |x-3| remains negative and the other part changes sign
-(x-3)>2x-1
.....x<4/3

x>3
both the parts of the equation are now positive
x-3>2x-1
......x>-2

taking common region from all the three equations.
Choice B

johanzac · Answer

Hi can someone help me understand the above solution (specifically on how each part within the modulus becomes negative or positive in each part pf the numberline)? 
I tried to solve this graphically on a numberline but, I couldnt arrive at this answer

Bunuel · Answer

Solve for x: |x - 3| > |2x - 1| A) x∈ (-∞, \frac{4}{3} ) ∪ (2, ∞) B) x∈ (-2, \frac{4}{3} ) C) x∈ (- \frac{4}{3} , 2) D) x∈ ( \frac{4}{3} , 2) E) x∈ (-∞, -2) ∪ ( \frac{4}{3} , ∞) APPORACH 1: Square the inequality to get rid of the modulus sign (note that we can safely do that since both sides of the inequality are non-negative): (|x - 3|)^2 > (|2x - 1|)^2 x^2 - 6x + 9 > 4x^2 - 4x + 1 3x^2 +2x -8 < 0 (x + 2)(x - 4/3) < 0 The "roots" are -2 and 4/3. The "<" sign indicates that the solution lies between the roots; therefore, the given inequality holds true for: -2 < x < 4/3. Answer: B. Check the links below. Solving Quadratic Inequalities - Graphic Approach: https://gmatclub.com/forum/solving-quad ... 70528.html Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm APPORACH 2: |x - 3| > |2x - 1| The critical points (aka key points or transition points) are 1/2, and 3 (the values of x for which the expressions in the absolute values become 0). So, we should consider the following three ranges: 1. x < 1/2 For this range, x - 3 < 0 and 2x - 1 < 0, so |x - 3| = -(x - 3) and |2x - 1| = -(2x - 1). Thus, for this range, we get -(x - 3) > -(2x - 1), which leads to x > -2. Hence, for this range, we get -2 < x < 1/2. 2. 1/2 ≤ x ≤ 3 For this range, x - 3 ≤ 0 and 2x - 1 ≥ 0, thus |x - 3| = -(x - 3) and |2x - 1| = 2x - 1. Thus, for this range, we get -(x - 3) > 2x - 1, which leads to x < 4/3. Hence, for this range, we get 1/2 ≤ x < 4/3. 2. x > 3 For this range, x - 3 > 0 and 2x - 1 > 0, thus |x - 3| = x - 3 and |2x - 1| = 2x - 1. Thus, for this range, we get x - 3 > 2x - 1, which leads to x < -2. Since the range we consider, x > 3, and the range we get, x < -2, have no overlap, then for the given range the inequality has no solution. So, we get that |x - 3| > |2x - 1| is true when -2 < x < 1/2 and 1/2 ≤ x < 4/3. When combined, these two ranges give -2 < x < 4/3. Answer: B. Hope it helps.

bumpbot · Answer

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Solve for x: |x-3|> |2x-1| A) x∈ (-∞, 4/3) ∪ (2, ∞)

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