Help with Pre-Cal Homework! ;)?

Question

SOLVE: (-z-3)/(z-3) + (11z+3)/(z^2-9) = (1-5z)/(z+3)

I tried doing this problem but after a few steps i get all mixed up.

Answer 1

It is very easy in a problem like this to get 'mixed up'.  I like to write the problem out evenly spaced across the entire page.  As you work the problem, keep sections lined up.  Now, I can't do any of this typing, but I'll do my best to help you.

(-z-3)/(z-3)  + (11z+3)/(z^2-9)        = (1-5z)/(z+3) Simplify where possible

-(z+3)/(z-3) + (11z+3)/[(z-3)(z+3)] = (1-5z)/(z+3) Get rid of fractions by multiplying by (z-3)(z+3)

-(z+3)(z+3) + (11z+3) = -(5z-1)(z-3) Multiply out and combine like terms

-(z²+6z+9) + 11z + 3 = -(5z²-16z+3)

-z² - 6z - 9 + 11z + 3 = -5z² +16z-3 Set equation equal to 0 - group together so not to confuse + or -

(-z² + 5z²) + (-6z + 11z - 16z) + (-9 +3 +3) = 0

4z² -11z -3 = 0

(z-3)(4z+1) = 0

z = 3 or z = -1/4

Check by plugging back in ... sorry, I don't have time to do that right now.  Please do it for yourself to make sure the numbers are correct :)

Answer 2

(-z-3)/(z-3) + (11z + 3)/(z^2-9) = (1-5z)/(z+3) :  factor the denominator in the second term.

-(z+3)/(z-3) + (11z + 3)/(z+3)(z-3) = -(5z+1)/(z+3): now find a common denominator and rewrite the numerator on the right side.

-(z+3)(z+3)/(z-3)(z+3) + (11z+3)/(z+3)(z-3) = -(5z+1)(z-3)/(z+3)(z-3): now add the numberators on the left and subtract the numberator on the right.

[-(z+3)(z+3) + (11z+3) +(5z+1)(z-3)]/(z+3)(z-3) = 0: Now multiply the terms on the left that need to be multiplied.

-(z^2 + 6z + 6) + (11z + 3) + (5z^2 -14z-3)/ (z+3)(z-3) = 0  Now distribute whatever you need to on the left:

{-z^2 - 6z + 6 + 11z + 3 + 5z^2 - 14z - 3}/(z+3)(z-3) = 0

Add like terms:

4z^2 -9z +6/(z+3)(z-3)

Now factor the numberator if possible:

(4z-3)(z+3)/(z+3)(z-3) = 0

Simplify the expression by cancelling out like terms:

(4z-3)/(z-3) = 0

Then 4z-3 = 0 since the denominator cannot equal zero:

4z = 3

z = 3/4  Done!