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Statement 1: x = 2 For (x+3) >= 0 x >= -3 [Not a solution as it doesn't satisfies the condition] x+3=4x-3 x = 2

For (x+3) < 0 x < -3 [Not a solution as it doesn't satisfies the condition] -(x+3) = 4x-3 x = 0 [Not a solution as it doesn't satisfies the condition]

Statement 2: x = 2 For (x+1) >= 0 x >= -1 [Not a solution as it doesn't satisfies the condition] x+1 = 2x-1 x = 2

For (x+1) < 0 x < -1 [Not a solution as it doesn't satisfies the condition] -(x+1) = 2x - 1 x = 0 [Not a solution as it doesn't satisfies the condition]

To save time, we should ignore all -ve cases of x as LHS is +ve and equal to RHS. So, for RHS to be positive x must be +ve.
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stmt1 ... when |x+3|<0 -(x+3) = 4x-3 5x=0,x=0 ....

Here is a typical mistake with moduli. You forgot to check your answer. It is possible to use -(x+3) only if |x+3|<0 ! But at x=0, |0+3| > 0.
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Stmt 1:|x + 3| = 4x – 3 Two cases: a. x+3=4x-3 ,x=0 b. -(x+3)=4x-3 ,x=0.

Both cases gives us x=0 hence SUFF.

Stmt 2:|x + 1| = 2x – 1 Again two cases: a.x+1=2x-1 x-2x=-1-1 x=2

b. -(x+1)=2x-1 x=0 Both case gives us diff. values hence insuff. IMO: A

Please point out my flaw.

First you made a mistake in calculation: see the red part, solution there is x=2. Second, which is more important the way of solving, from my point of view, is incorrect.

Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient.

(2) \(|x+1|=2x-1\) --> the same here --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

So you see that you don't even need to find exact value(s) of \(x\) to answer the question.

Stmt (1) First we squire both sides of the equality: |x + 3| = 4x – 3 x^2+6x+9=16x^2-24x+9 15x^2-30x=0 divide both sides to 15 and we have: x^2-2x=0 x(x-2)=0 x=0 but x=0 is not a root of the equation since |0 + 3| is not equal to 4*0 – 3= -3 because absolute value cannot equal to digit below zero x-2=0 x=2 is a root of the equality, since |2 + 3| = 4*2 – 3=> 5 both sides hence (1) is sufficient

stmt (2) |x + 1| = 2x – 1 squire both sides again and we have: x^2+2x+1=4x^2-4x+1 3x^2-6x=0, divide both sides to 3 and have: X^2-2x=0 x(x-2)=0 x=0 but x=0 is not a root of the equation since |0 + 1| is not equal to 2*0 – 1= -1 because absolute value cannot equal to digit below zero x-2=0 x=2 is a root of the equality, since |2 + 1| = 2*2 – 1=> 3 both sides

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(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient.

(2) \(|x+1|=2x-1\) --> the same here --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

So you see that you don't even need to find exact value(s) of \(x\) to answer the question.

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