Basic algebra question

\(\frac{2}{x-3} < = 2\)

Reciprocal both sides

\(\frac{x-3}{2} > = \frac{1}{2}\)

With denominators same, numerator can be equated

x > = 4

Let's break this equation into two:

a) \(\frac{2}{{x - 3}} < 2\) &

b) \(\frac{2}{{x - 3}} = 2\)

Equation (b) can be simplified as
\(2 = 2 (x - 3)\)

\((x - 3) = 1\)

So, \(x = 4\)

For equation (a), we will need to consider two cases:
Case 1: \(x - 3\) is positive or \(x - 3 > 0\) or \(x > 3\)

In this case, we can multiply both the sides of the inequality by \(x - 3\) without affecting the sign. So, we have

\(2 < 2 (x - 3)\)

\((x - 3) > 1\)

So, \(x > 4\)

Now, from our assumption we have \(x > 3\). We will need to take the more restrictive condition. So, we get \(x > 4\).

Case 2: \(x - 3\) is negative or \(x - 3 < 0\) or \(x < 3\)

In this case, when we multiply both the sides of the inequality by \(x - 3\), the sign will reverse. So, we have

\(2 > 2 (x - 3)\)

\((x - 3) < 1\)

So, \(x < 4\)

Again, from our assumption we have \(x < 3\). As before, we will need to take the more restrictive condition. So, we get \(x < 3\).

Combining all the conditions we get \(x \ge 4\) or \(x < 3\).

Hope that makes it clear

Follow these steps:

Simply reciprocal both the sides

Since denominators are same, they can be cancelled

Then normally solve for inequality which gives the following solution:

x >=4

[quote="danzig"]Please, your help. If there is an algebraic method to solve this question without picking numbers? Thanks!


\(\frac{2}{(x-3)}= \frac{1}{2}\)

Put x = 1. The second inequality does not hold. The reason for this is that when you take the reciprocal, you are in effect, multiplying both sides by (x -3) and then dividing both sides by 2.
If (x -3) is negative, you cannot multiply both sides by it without flipping the sign. Hence it is not correct. If you take the reciprocal, you are assuming that x>3 and hence losing out on a range of values that x can take. Instead do this:

\(\frac{2}{(x-3)}= 0\)

This gives us x >= 4 or x < 3

If you are unclear how we got this, check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... e-factors/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... ns-part-i/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... s-part-ii/
and comments of the third post.