Can someone hlep me solve these 2 inequalities?

Question

1. ((x-2)/x) + (2/(x-1)) ≤ 2

2. (x+12)(x-9)/(x-1)≥0

Answer 1

Before I get started, remember to *never* multiply or divide by a variable in an inequality, because this eliminates critical values. With that said ...

1) (x - 2)/x + 2/(x - 1) ≤ 2

First, merge the two fractions into one, by giving them a common denominator.

[ (x - 2)(x - 1) + 2(x) ] / [ x(x - 1) ] ≤ 2

Simplify.

[ x^2 - 3x + 2 + 2x ] / [ x(x - 1) ] ≤ 2

[ x^2 - x + 2 ] / [ x(x - 1) ] ≤ 2

Now, move the 2 to the left hand side and do the exact same thing.

[ x^2 - x + 2 ] / [ x(x - 1) ] - 2 ≤ 0

[ x^2 - x + 2 ] / [ x(x - 1) ] - 2x(x - 1)/[x(x - 1)] ≤ 0

[ x^2 - x + 2 - 2x(x - 1) ] / [ x(x - 1) ] ≤ 0

[ x^2 - x + 2 - 2x^2 + 2x ] / [ x(x - 1) ] ≤ 0

[ -x^2 + x + 2 ] / [ x(x - 1) ] ≤ 0

[ (-1)(x^2 - x - 2) ] / [ x(x - 1) ] ≤ 0

[ (-1)(x - 2)(x + 1) ] / [ x(x - 1) ] ≤ 0

Multiplying both sides by (-1) flips the inequality sign.

[ (x - 2)(x + 1) ] / [ x(x - 1) ] ≥ 0

Here is where we obtain our critical values; critical values are defined to be what makes the left hand side equal to 0, or what makes it undefined. I'm going to put an asterisk (*) beside the critical values that make the left hand side undefined.

By inspection, our critical values are 2, -1, 0*, 1*.
Since we have greater than _or equal to_, we have to include the critical values in our solution set which are defined. We exclude the values which are not defined.

At this point, make a number line consisting of the critical points:

. . . . . . . (-1) . . . . . . . (0) . . . . . . . . . (1) . . . . . . . . . (2) . . .

Now, we *test* each region around the critical values. All we have to do is test a single value in each region, and apply that value to the left hand side of our inequality, [ (x - 2)(x + 1) ] / [ x(x - 1) ]. All we care about is whether that value is positive or negative.

Test x = -2:
[ (x - 2)(x + 1) ] / [ x(x - 1) ] = [ (neg) (neg) ] / [ (neg)(neg) ]
= positive.
Mark the region as positive.

. . .{+} . . (-1) . . . . . . . (0) . . . . . . . . . (1) . . . . . . . . . (2) . . .

Test x = -1/2.
[ (x - 2)(x + 1) ] / [ x(x - 1) ] = [ (neg)(pos) ] / [ (neg)(neg) ]
= negative.
Mark the region as negative.

. . .{+} . . (-1) . . . {-} . . (0) . . . . . . . . . (1) . . . . . . . . . (2) . . .

I'm going to skip the details of testing the other regions.

. . .{+} . . (-1) . . . {-} . . (0) . . . {+}. . . . (1) . . . {-} . . . (2) . {+} ...

Since we are interested in the regions which are greater than or equal to 0 (as per the question), we are interested in the positive regions. Remember to include the defined critical values with a square bracket and exclude the undefined ones with a round bracket. As a reminder, our critical values are
2, -1, 0*, 1*.

In interval notation, our solution set is

(-infinity, -1] U (0, 1) U [2, infinity)

{negative infinity and infinity always have round brackets, and as you can see, we appropriately bracketed the values in our solution set based on what was undefined}

2) (x + 12)(x - 9)/(x - 1) ≥ 0

Critical values: -12, 9, 1*

Number line test:

. . . . . . . . (-12) . . . . . . . . . (1*) . . . . . . . . . . (9) . . . . . . . . . .

Test x = -13;
(x + 12)(x - 9)/(x - 1) = (neg)(neg)/(neg) = negative.

. . . {-}. . . (-12) . . . . . . . . . (1*) . . . . . . . . . . (9) . . . . . . . . . .

Test x = 0:
(x + 12)(x - 9)/(x - 1) = (pos)(neg)/(neg) = positive.

. . . {-}. . . (-12) . . . {+} . . . . (1*) . . . . . . . . . . (9) . . . . . . . . . .

Skipping the details of the rest and completing the number line:

. . . {-}. . . (-12) . . . {+} . . . . (1*) . . . . {-}. . . . . . (9) . . . . {+} . . . .

We are interested in positive regions. Our answer in interval notation is

[-12, 1) U [9, infinity)

*******
As a last note, you'll notice that in both of these cases, we get an alternating pattern of positive and negative. As long as you don't have any even-powered factors, this is what's going to happen. If we take another example like this one:

(x - 2)^2 (x + 1) / (x - 5) ≤ 0

Then we won't get alternating signs; around the double factor, we will get the same sign, and the alternating pattern will commence.

. . . . . . . . (-1) . . . . . . . . (2) . . . . . . . . (5*) . . . . . . . . .

This time I'm going to test one region.
Test x = -2:
(x - 2)^2 (x + 1) / (x - 5) = (pos)(neg) / (neg) = positive.

. . . {+} . . . (-1) . . . . . . . . (2) . . . . . . . . (5*) . . . . . . . . .

Alternate signs except at the double factor at x = 2.

. . . {+} . . . (-1) . . . {-} . . . (2) . . . {-} . . . (5*) . . . .{+}. . . . .

We want the negative regions (≤ 0 as per the question). Our solution set is

[-1, 2] U [2, 5)

Answer 2

2)

(x+12)(x-9)/(x-1) ≥ 0

Please follow me step by step. I don't want just to solve the problem, I want to help you to understand how to solve it.

Solution:

A) x=1 is surely not a solution. (OK?)

B) Suppose that x < 1 => x-1 < 0 then (x+12)(x-9) ≤ 0.
B1) Suppose (x+12) ≤ 0 and (x-9) ≥ 0. That's impossible. (OK?)
B2) Suppose (x+12) ≥ 0 and (x-9) ≤ 0 then x ≥ -12 and simultaneously x ≤ 9 => -12 ≤ x ≤ 9.

Including presupposition B) we get the first branch of the solution: -12 ≤ x ≤ 1 (OK?)

C) Suppose that x > 1 => x-1 > 0 then (x+12)(x-9) ≥ 0.
C1) Suppose (x+12) ≤ 0 and (x-9) ≤ 0 then x ≤ -12 and simultaneously x ≤ 9 => x ≤ -12. That's impossible. (OK?)
C2) Suppose (x+12) ≥ 0 and (x-9) ≥ 0 then x ≥ -12 and simultaneously x ≥ 9 => x ≥ 9.

Including presupposition c) we get the second branch of the solution: x ≥ 9 (OK?)

The final solution is:

-12 ≤ x ≤ 1 and x ≥ 9

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