Resolve [x - 1]/x + 3 > 2 into partial fractions?

Answer 1

Assuming (x+3) is all in the denominator, you multiply by (x+3). However, you have to flip the inequality if you multiplied by a negative number, so this splits the problem into two cases (x+3 is negative, x+3 is positive).

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case 1: x+3 is positive:

(x - 1)/(x + 3) > 2
x - 1 > 2x + 6
-7 > x
x < -7

However, x+3 is positive is also required;

x + 3 > 0
x > -3

There are no numbers that are both less than -7 and greater than -3, so this case yields no solution.

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case 2: x+3 is negative:

(x - 1)/(x + 3) > 2
x - 1 < 2x + 6 (note flipped inequality!)
-7 < x
x > -7

However, x+3 is negative is also required;

x + 3 < 0
x < -3

Combined, this gives us the solution -7 < x < -3. Since the first case yielded no solution, the solution to this second case is the full solution to the problem.

That's the answer: x must be between -7 and -3.

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Now let's check the answer. First, let's pick -5 which is in our range:

(x - 1)/(x + 3) >? 2
(-5 -1)/(-5 + 3) >? 2
-6 / -2 >? 2
3 > 2

Now let's try -10 which is outside of it:

(x - 1)/(x + 3) >? 2
(-10 - 1)/(-10 + 3) >? 2
-11 / -7 >? 2
11 / 7 >? 2

11/7 is not greater than 2, so we have correctly excluded it.

And let's try -1, which is also not in our range:

(x - 1)/(x + 3) >? 2
(-1 -1)/(-1 + 3) >? 2
-2 / 2 >? 2
-1 >? 2

-1 is not greater than 2, which is also legitimately excluded.

Answer 2

x-1>2x+6
x-2x>6+1
-x>7
x>-7

Answer 3

sara...
when u shift the -ve over shoudn't u change the position of the sign?
x>-7 <-----instead of this...
shouldnt it be
x<-7 ??