how do you solve... (x+2)/(x-1) is less than or equal to 0?

Answer 1

(x + 2)/(x - 1) <= 0

To solve this, you must first determine your critical values. This is done by solving the corresponding equation
(x + 2)/(x - 1) = 0

Critical values are what make (x + 2)/(x - 1) equal to 0 or what makes it undefined. In this case, -2 makes it 0, and 1 makes it undefined. So your critical values are
x = {-2, 1*}

[Note: I put a * beside the 1 to note that it is the value that makes it undefined.]

Now, make a number line consisting of your values.

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

We test each region around the critical number by plugging in a *single* value within that region for positivity or negativity. We want to test (x + 2)/(x - 1).

Test -10: (x + 2)/(x - 1) = (-10 + 2)/(-10 - 1) = (-8)/(-11), and we only care that it's positive. Mark the region as positive.

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

Test x = 0. (x + 2)/(x - 1) = (2)/(-1) = negative. Mark the region as negative.

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

Test x = 10. You can test it out for yourself; it will be positive.

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

Now, look back to what the question is asking for; it is asking for "less than _or equal to_ zero". Because of the "or equal to" clause, not only do you want the regions which are negative, but you want to *include* the critical value that made the left hand side equal to 0. This would be (-2). You also want to EXCLUDE the value that made the function undefined (hence the *). Therefore, in interval notation, our solution is

x E [-2, 1) {Note the round bracket to exclude the 1, and the square bracket to include the -2}

In set notation, it is

{x | -2 <= x < 1, x E R}

"The set of all x such that x is greater than or equal to -2 and x is less than 1, for all real numbers x."

***
As a side note, it may be tempting to multiply both sides of the inequality by (x - 1); that is

(x + 2)/(x - 1) <= 0
(x + 2) <= 0

This would be a *wrong* step because it would be eliminating values. Be careful with inequalities in this regard.

Answer 2

5cosx = 2sin^2 + 1 sin^2x + cos^2x = 1 <= Pythagorean identity 5cosx = 2(1-cos^2x) + 1 5cosx = 3 - cos^2x 2cos^2x + 5cosx -3 = 0 Substitute a = cosx, which reduces this to a standard quadratic: 2a^2 + 5a - 3 = 0 Factor (by whatever method you choose) (a+3)(2a-1) = 0 => a+3 = 0 OR 2a-1 = 0 Which implies that: cosx + 3 = 0 OR cosx = 1/2 The first is clearly impossible, so we only need to look at the second. The solutions to this are pi/3 and 5pi/3

Answer 3

x equals any negative number, just plug any negative number in and you will get 0 or less than 0

Answer 4

x < +1

It can not = 1 or denom = 0 = undefined