Inequalities: how to solve 1/x + 1 > 2 (i got x = -1/2)

[kazafz]

Hi the question that i want to ask is -

How to solve the following inequality:
1/x+1 > 2

For this question the thing i did was move the "x + 1" to the other side so it becomes 1 > 2(x + 1) which is 1 > 2x+2 then i got x = -1/2 but in the answer it had
-1 < x < -1/2 and the problem is i don't know how to get the -1 from the equation. Can someone teach me please.

Thanks

 

[tkhunny]

Re: Inequalities

1) Please fix your notation. 1/x+1 is NOT the same as 1/(x+1).

2) How did you get an Equality for your answer? Where did the INequality go?

3) "move" is not an appropriate operation. If you had realized that you were using multiplication, perhaps you would have recalled that multiplying by negative numbers requires the reversal of the inequality. In this case, x+1, you do not know if it is positive or negative, so you must be very careful.

Assume x+1 is positive. x+1 > 0 or x > -1. Then solve as you wish. 1 > 2(x+1)

Assume x+1 is negative. x+1 < 0 or x < -1. Then solve as you wish. 1 < 2(x+1)

x+1 cannot be zero. Do you see why?

 

[kazafz]

Re: Inequalities

1) Please fix your notation. 1/x+1 is NOT the same as 1/(x+1).

2) How did you get an Equality for your answer? Where did the INequality go?

3) "move" is not an appropriate operation. If you had realized that you were using multiplication, perhaps you would have recalled that multiplying by negative numbers requires the reversal of the inequality. In this case, x+1, you do not know if it is positive or negative, so you must be very careful.

Assume x+1 is positive. x+1 > 0 or x > -1. Then solve as you wish. 1 > 2(x+1)

Assume x+1 is negative. x+1 < 0 or x < -1. Then solve as you wish. 1 < 2(x+1)

x+1 cannot be zero. Do you see why?

Yes I see why x + 1 cannot be 0 because if it was zero then the answer would be undefined. So does these positive and negative cases occur for questions when theres "something" over x? For these kinds of questions to solve them do i need to first find a number which will make answer undefined first so i know that number is not true then i find the other one?

 

[skeeter]

\(\displaystyle \frac{1}{x+1} > 2\)

\(\displaystyle \frac{1}{x+1} - 2 > 0\)

\(\displaystyle \frac{1}{x+1} - \frac{2(x+1)}{x+1} > 0\)

\(\displaystyle \frac{1 - 2(x+1)}{x+1} > 0\)

\(\displaystyle \frac{-2x - 1}{x+1} > 0\)

\(\displaystyle \frac{2x + 1}{x+1} < 0\)

critical values for this inequality are where the rational function on the left side of the inequality equals 0 or is undefined, namely at x = -1/2 and x = -1. These two x-values break the possible values for x into three regions ...

x < -1

-1 < x < -1/2

x > -1/2

pick a single value for x from each region and substitute into the inequality ... if that value of x makes the inequality true, then all values of x in that region will make the inequality true. if the value of x makes the inequality false, then all values of x in that region make the inequality false.

the solution set for the inequality are those values of x that make the inequality true.

for this particular inequality, the solution set is -1 < x < -1/2

 

[Deleted member 4993]

Then check your answer by graphing the function in your calculator (or computer - or by hand).