Bunuel wrote:
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.
Answer: D.
Hope it helps.
For those more inclined to solving equations, here is a different approach:
(1) |x+3|=4x-3
If x+3>0 then:
a) x > -3 AND
b) x+3 = 4x-3 => 3x=6 => x=2.
Since x=2 does NOT contradict x>-3, x=2 is a solution and x is positive
However if x+3<0 then:
a) x<-3 AND
b) x+3=-4x+3 => x=0. BUT x must be less than -3, so x=0 is NOT a solution.
Therefore only the case where x+3>0 can be possible being x=2 the only possible solution, therefore always positive.
(2) |x+1|=2x-1 Works exactly the same:
If x+1>0 then:
a) x > -1 AND
b) x+1 = 2x-1 => x = 2
Since x=2 does NOT contradict x>-1, x=2 is a solution and x is positive
However if x+1<0 then:
a) x<-1 AND
b) x+1 = -2x+1 => x = 0. BUT x must be less than -1, so x=0 is NOT a solution.
Therefore only the case where x+1>0 can be possible being x=2 the only possible solution, therefore always positive.
In BOTH cases the only possible value for x is always positive. Therefore D is the answer.