Rational graphs; solving an inequality

Hi there i have the question

\frac{3x + 4}{x - 2} > 1

I have worked out that one of the solutions is x = -3, so this is x < -3, but I don't know how to find the other solution, which is given in the answers as x > 2.

Please could someone explain how to find this solution.

I will admit I have been lazy and haven't sketched the graph.

Thanks
:smile:

(edited 12 years ago)

Sketch the graph.

Original post by jackie11

Hi there i have the question

\frac{3x + 4}{x - 2} > 1

I have worked out that one of the solutions is x = -3, so this is x < -3, but I don't know how to find the other solution, which is given in the answers as x > 2.

Please could someone explain how to find this solution.

I will admit I have been lazy and haven't sketched the graph.

Thanks
:smile:

Always start by moving all the terms to the LHS, so that you have 0 on the RHS.

Then put everything over a common denominator and factorise fully.

You then have three choices:

(1) graph the function
(2) make a table of critical values - where one of the factors is zero. Work out ranges of x between the critical values and which brackets are +ve/-ve.
(3) Consider two cases - where the denominator is -ve and where it's +ve.

You can also multiply through by the denominator squared, as has been suggested, but you just end up with the same brackets as method (2).

Reply 3

Hasufel

16

you can multiply both sides of the inequality by (x-2)^2 - because it`s not negative, and collect terms on the LHS, to get... (x-2)(x-3)>0.......then sketch (x-2)(x-3) = y, solve it for =0 and examine where the graph is +ve...

(edited 12 years ago)

Reply 4

Refrigerator

3

Original post by Hasufel

you can multiply both sides of the inequality by (x-2)^2 - because it`s not negative, and collect terms on the LHS, to get... (x-2)(x-3)>0.......

not quite, when x=2 in the original inequality the value is undefined since you'd be dividing by zero so there will be a gap in the g

Reply 5

Hasufel

16

you`re forgetting, the inequlity you get is >0, not =0, since to begin with the rational inequality is >1, which means the denominator can`t possibly be zero to start with (otherwise the question wouldn`t make sense unless x=-4/3 [which]) even looking at the graph of 3x+4/x-2 -1 =y, my original quadratic holds, since, fair enough graph is undefined at 2 , but who cares what it`s doing AT 2? it is +ve as it goes past -3 (to minus infinity) and likewise after 2.

Ziishi?

(edited 12 years ago)

Reply 6

f1mad

13

Take the 1 over and get a common denominator:

3x+4/x-2 - (x-2)/x-2>0

Now combine the fractions then work out the critical values.

Then use a sign diagram to find the -'s and +'s.

Since its >0 you will want to identify the +'s here.

Reply 7

f1mad

13

Original post by f1mad

Take the 1 over and get a common denominator:

3x+4/x-2 - (x-2)/x-2>0

Now combine the fractions then work out the critical values.

Then use a sign diagram to find the -'s and +'s.

Since its >0 you will want to identify the +'s here.

Lol who negged?

My solution is correct.

Problem?

Reply 8

jackie11

OP

4

ok thank you all for your input, its very appreciated!!! :smile:

Reply 9

TenOfThem

18

Consider when the denominators positive and the denominator is negative as 2 separate scenarios

Spoiler

Is my preferred method

Reply 10

Hasufel

16

Original post by TenOfThem

Consider when the denominators positive and the denominator is negative as 2 separate scenarios

Spoiler

Is my preferred method

that`s a good one too - a more "mathematical" way of doing it! (and my prefered one too - combined with, when you have lots of bracketed expressions, then a sign diagram is handy))

(edited 12 years ago)

Reply 11

jackie11

OP

4

ok thanks

Rational graphs; solving an inequality

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