SOLUTION: (3/(2z+1))-(z/(z-2))=((-3z^2+2z=58)/(2z^2-3z-2))

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Question 214127: (3/(2z+1))-(z/(z-2))=((-3z^2+2z=58)/(2z^2-3z-2))
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming the second equals sign is supposed to be a plus sign (since plus is shift = and since it make the problem solvable) making your problem:
%283%2F%282z%2B1%29%29-%28z%2F%28z-2%29%29=%28%28-3z%5E2%2B2z%2B58%29%2F%282z%5E2-3z-2%29%29
To avoid mistakes commonly made with subtractions, we'll rewrite the left side as an addition before proceeding:

Solving this, or any equation with fractions, is easier if you start by eliminating the fractions. And the easiest way to eliminate the fractions is:
  1. Find the lowest common denominator (LCD) of all the fractions on both sides of the equation. To find the LCD of the fractions:
    1. Factor each denominator
    2. The LCD is the product of the highest power of each different factor.
  2. Multiply both sides of the equation by the LCD.

So we'll start by factoring the denominators. With the two denominators on the left side the only factoring that can be done is to factor out a 1. The denominator on the right is a trinomial which can be factored:

Now that the denominators are factored we can see that there are 2 distinct factors (3 if you count the 1's): (2z+1) and (z-2). And the highest exponent on each is just 1. So the LCD is simply: (2z+1)(z-2).
The next step is to multiply both sides of the equation by the LCD:

Using the distributive property on the left side we get:

Now we can see the "magic of the LCD" as all the denominators cancel out!

leaving:
%28z-2%293+%2B+%282z%2B1%29%28-z%29+=+-3z%5E2+%2B+2z+%2B+58
and the fractions are gone!! Now we multiply out the left side (again with the Distributive Property):
3z+-+6+%2B+%28-2z%5E2%29+%2B+%28-z%29+=+-3z%5E2+%2B+2z+%2B+58
Adding the like terms we get:
%28-2z%5E2%29+%2B+2z+-+6+=+-3z%5E2+%2B+2z+%2B+58
Since we have a quadratic equation we'll gather everything to one side by adding the opposite of each term on the right to both sides giving:
z%5E2+-+64+=+0
Now we solve this by factoring (or, if you prefer, using the quadratic formula). Since the left side is a difference of squares it is easily factored:
%28z%2B8%29%28z-8%29+=+0
If this product is zero, then one of the factors must be zero:
z+8 = 0 or z-8 = 0
Solving each of these we get:
z = -8 or z = 8.
As a general rule, when you solve equations with variables in the denominator (like this one), you should either check your answers or at least check to make sure the answers do not make any denominators zero. If a possible solution makes a denominator zero then we have to discard that answer.
If we just check the denominators it is easiest to check the factored form of the denominators. So we'll check to see if z=-8 or z=8 make any of the following zero:
(2z+1), (z-2) or (2z+1)(z-2)
I'll leave the rest of the checking up to you. But you should find that neither solution makes any of the denominators zero. So we do not have to discard either solution.
So your problem has two solutions: z = -8 or z = 8.