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31 Aug 2014, 12:46
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Please, your help. If there is an algebraic method to solve this question without picking numbers? Thanks!
\(2/(x-3)\)\(<= 2\)
\(2/(x-3)\)\(<= 2\)
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31 Aug 2014, 23:55
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danzig
\(\frac{2}{x-3} < = 2\)
Reciprocal both sides
\(\frac{x-3}{2} > = \frac{1}{2}\)
With denominators same, numerator can be equated
x > = 4
01 Sep 2014, 04:07
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danzig
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
