Question

I feel so helpless....

Answer 1

\[\frac{ 5 }{ x-3 }\ge\frac{ 4 }{ x-2 }\]

Answer 2

@amistre64 @ganeshie8 @phi @ParthKohli @thomaster

Answer 3

@Australopithecus

Answer 4

why is it \[\frac{ 5 }{ x-3 }-\frac{ 4 }{ x-2 }\ge0\]

Answer 5

instead of \[\frac{ 5 }{ x-3 }-\frac{ 4 }{ x+2 }\ge0\]

Answer 6

because there is something called the addition/subtraction property of inequality which means that adding or subtracting to both sides of an inequality does not affect the nature of the inequality. When subtract \[\frac{ 4 }{ x-2 }\] the result stays the same on both sides
\[\frac{ 4 }{ x+2 }\] is a different fraction altogether. Consequently when you subtract the first fraction from both sides, the first fraction does not change into the second fraction.

Answer 7

can you keep solving the problem

Answer 8

\[\frac{ 5(x-2) }{ (x-3)(x-2) }-\frac{ 4(x-3) }{ (x-3)(x-2) }\]

Answer 9

why does it turn into this? where do the other (x-3)(x-2) go? \[\frac{ 5(x-2)-4(x-3) }{ (x-3)(x-2) }\ge0\]

Answer 10

LCM

Answer 11

?

Answer 12

you take the LCM.

Answer 13

can you elaborate

Answer 14

like
\[\frac{ 1 }{ 2 }+\frac{ 1 }{ 2 }=\frac{ 2 }{ 2 } = 1\]

Answer 15

yea but in this case

Answer 16

\[\large
\frac ab - \frac cd = \frac{ad-bc}{bd}
\]

Answer 17

i know that but shouldnt it be more? like in the denominator shouldnt it be (x-3) (x-3) (x-2)(x-2)

Answer 18

No. The least COMMON denominator is (x-3)(x-2)

Answer 19

hmmm okay gotcha

Answer 20

i still dont understand why bringing it to the other side doesn't make it a (x+2) instead of a (x-2)

Answer 21

There is a negative sign in front of the entire fraction. \(\Large \frac{4}{x-2}\) on the right hand side becomes \(\Large -\frac{4}{x-2}\) on the right hand side.

Answer 22

I meant on the left hand side.

Answer 23

but why doesn't it become a negative x+2

Answer 24

hmmm okay and theres no other instances where its not like this right?

Answer 25

You are changing the sign on just one part of a fraction rather than the whole fraction which is wrong. It is either \(\Large -\frac{4}{x-2}\) or \(\Large +\frac{4}{2-x}\)

Answer 26

No. 50 / (20 - 10) = 5
If you take it to the other side it becomes either - 50 / (20 - 10) = -5 OR
+ 50 / (10 - 20) = 50 / (-10) = -5.

But if you change the sign on just a part of the fraction then it becomes: 50 / (20 + 10) = 5/3 2hich is wrong.

Answer 27

wait what? that last part...

Answer 28

You were asking why 4/(x-2) when taken to the other side does not become -4/(x+2). If I take 50 / (20 - 10) to the other side it is NOT -50 / (20 + 10) which is -5/3. It should be -50/(20-10) = -5.

Answer 29

okay gotcha well can you help me solve the rest of this problem?

Answer 30

\[
\frac{ 5 }{ x-3 }\ge\frac{ 4 }{ x-2 } \\
\text{Subtract } ~ \frac{ 4 }{ x-2 } ~~\text{from both sides: } \\
\frac{ 5 }{ x-3 }-\frac{ 4 }{ x-2 } \ge 0\\
\frac{ 5(x-2) }{ (x-3)(x-2) }-\frac{ 4(x-3) }{ (x-2)(x-3) } \ge 0\\
\]

Answer 31

\[
\frac{ 5(x-2) }{ (x-3)(x-2) }-\frac{ 4(x-3) }{ (x-2)(x-3) } \ge 0\\
\frac{ 5(x-2)-4(x-3) }{ (x-3)(x-2) } \ge 0\\
\frac{ 5x-10-4x+12) }{ (x-3)(x-2) } \ge 0\\
\frac{ x+2 }{ (x-3)(x-2) } \ge 0\\
\]

Answer 32

\[\frac{ 5x-10-4x+12 }{ (x-3)(x-2) }\]

Answer 33

okay got it

Answer 34

x=-2 x=3 x=2

Answer 35

\[
\frac{ x+2 }{ (x-3)(x-2) } \ge 0\\
\]

The important values of x here are the ones that will make the numerator/denominator zero. They are x = -2, 2 and 3.
Pick a suitable point in the interval (-infinity, -2); (-2, 2); (2, 3) and (3, +infinity) and see in which intervals it is >= 0.

Answer 36

do you always have to plug in random numbers to check your answer?

Answer 37

It is an easy method to quickly find the solution.
In the interval (-infinity, -2), let us pick -3. (-3+2) / { (-3-3)*(-3-2) }
negative / positive = negative. So LHS is not >= 0 here.

Answer 38

You don't have to actually find the number just the sign will do.

Answer 39

For (-2, 2) pick x = 0.

Answer 40

wait is it > or the greater than or equal to sign

Answer 41

The original problem had >= (greater than or equal to sign) and so it will continue to remain as greater than or equal to sign throughout.

Answer 42

oh okay is that what you mean by ">"

Answer 43

So we ruled out (-infinity, -2).
The interval (-2, 2) is a solution. But since it is "greater than OR Equal to zero" we have to INCLUDE -2 in the solution and so the interval becomes [-2, 2).

Try the other two intervals too.

Answer 44

The open bracket [ includes the point -2, but the open parenthesis ( excludes the point -2.

Answer 45

okay I'm gonna be honest with you.... this part is the only part i dont understand.... how do you know what to put? Like what am i trying out?

Answer 46

i understand the meanings of the brackets but like I don't understand what I'm putting in it...

Answer 47

Looking at the expression we figured what values make the numerator or the denominator zero. Then we arranged those x values in increasing order. They were -2, 2 and 3.