Advanced Math/Inequality
Expert: Sherry Wallin - 12/7/2008
QuestionThe problem I was given was to solve for "x" algebraically when
(x+5) divided by (x-3) is greater than or equal to zero.
I know that the denominator cannot equal zero, therefore x must be greater than 3, but beyond that I have no idea how to approach the problem, or what steps to take to solve for all values of x
Answer
Hi Steph~
This is a problem with reasoning and math. What do you know about when a number (a rational number, a fraction) = 0? Right, when the numerator is 0, correct? And what do you know about the sum of two numbers over a denominator? Isn't 1/7 + 3/7 = (1+3)/7? and vice versa?
So write (x+5)/(x-3) as x/(x-3) + 5/(x-3)>= 0
So x/(x-3)>= -5/(x-3)
Now use cross multiplication to get rid of the denominators.
x(x-3) >= -5(x-3)
x^2-3x >=-5x + 15
x^2-3x +5x - 15 >= 0
x^2 +2x-15 >= 0
(x+5)(x-3)>=0 remembering as you stated above that x != 3
so possible solutions are x >= -5 and x >= 3. But you can not include 3 so it is x>3. So take your "critical values" and put them on a number line and check those values in your factors. [Please see image in Word document attached]. You don't need to get actual values just if they are positive or negative or 0 since you are looking for all numbers >= 0.(x-3)(x+5)
So my "check points" are -6 -> (-)(-)= +
1: -> (-)(+) = -
4: -> (+)(+) = +
So the intervals that our rational expression is >= 0 is
(- infinity,-5]union(3,infinity). Notice the square bracket next to the -5 to indicate the solution could be -5 and the curved parenthesis next to 3 to indicate you can't include 3.
I hope this clearly helps you solve the problem and gain new understanding.
Math Prof