Inequality questions

When you solve a problem, you can replace it with equivalent problems and solve those. However, if you multiply both sides by x-2 the new problem is not going to be equivalent to the initial one because there are going to be two situations:

4 < 2*(x-2) if x-2 > 0
4 > 2*(x-2) if x-2 < 0

because when you multiply both sides by the same positive number, inequality sign doesn't change, and the sign changes when you multiply both sides by the same negative number (for example if (-2) < (-1) and you multiply both sides by (-1) it would be incorrect to say: (-1)*(-2) < (-1)*(-1) or 2 < 1)

In order to avoid this problem, I mutliply both sides by a positive number (x-2)^2 and keep the inequality sign unchanged. I hope that helps.

because x is a variable and we have an inequality here. u cant cross multiply unles u know the sign of the variable. a negative flips the inequality sign

here is my question: when you solve both of those inequalities, you get that either x is less than 4, or that x is greater than 4. how do you get to the final answer of x<2 and x>4 ?

I still have a confusion so am writing again...

I came across a ques:

Q-Z Find the value of x in (3x^2 + x) / (x^2 + 2) < 3.

A- here we did a normal cross multiply -
3x^2 + x < 3(x^2 + 2)
3x^2 + x < 3x^2 + 6
x < 6. answer (correct)

so whats the difference in that ques & (solution set of inequality 4/x-2 < 2 ) ??

cant we do -> 4 < 2(x-2)
4 < 2x - 4
8 < 2x
x > 4. (wrong)

If we can't then, why cross multiplication in Q-Z. moreover with what no u multiplied in 4/x-2 < 2. why not x-2 to both sides. ??

i think the difference here is that in Q-Z, you are dealing with x^2, and so those terms will always be positive no matter what.

thanks for the answer guys...

\(\frac{4}{(x-2)} < 2\)

my logic:

1. it is true for all negative (x-2) or for x<2
2. the function \(f=\frac{4}{(x-2)}\) is \(\infty\) at x close to 2 at the greater side.
3. When x increases from 2, the function f always decreases and equal to 2 at x=4. Therefore, f<2 at x>4
4. Combine two conditions: x<2 & x>4

how can you show this algebraically ?

im getting 4 >x....

1. the inequality is true when 4/(x-2) is negative.
2. 4/(x-2) is negative when (x-2) is negative
3. (x-2) is negative when x x<2)