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.