SOLUTION: So confused with rational expressions! Please help! Is this really as easy as it seems? I can't afford to fail another test! Having the hardest time with the LCM and the dividi

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Question 576908: So confused with rational expressions! Please help! Is this really as easy as it seems? I can't afford to fail another test!
Having the hardest time with the LCM and the dividing!
1) Subtract. Simplify if possible. (17cb/(c^2-b^2)) - ((c-b)/(c+b))
2) Add. Simplify if possible. (25f/(2f-10)) + (2f/(10f-50))
3) Divide and simplify. ((z^2-81)/z) / ((z-9)/(z+1))
4) Find the LCM of z^2+12z+36 and z^2+3z-18. Use Factored Form. I think the answer is (z+3), is this correct?
5) Find the LCM of 21y^7 and 147y^9. I think this is 3y^7, is this right?
6) Simplify by removing by factors of 1. 42y^5/54y^3 Is this 7y^2/9?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1) Subtract. Simplify if possible. (17cb/(c^2-b^2)) - ((c-b)/(c+b))
17cb%2F%28c%5E2-b%5E2%29-+%28c-b%29%2F%28c%2Bb%29=17cb%2F%28%28c+-+b%29%28c%2Bb%29%29-+%28c+-+b%29%2F%28c%2Bb%29=17cb%2F%28%28c+-+b%29%28c%2Bb%29%29 - %28c+-+b%29%28c-b%29%2F%28%28c%2Bb%29%28c-b%29%29=%2817cb-%28c-b%29%28c-b%29%29%2F%28%28c%2Bb%29%28c-b%29%29=%2817cb-%28c%5E2-2bc%2Bb%5E2%29%29+%2F%28%28c%2Bb%29%28c-b%29%29=%2819cb-c%5E2-b%5E2%29%2F%28%28c%2Bb%29%28c-b%29%29
2) Add. Simplify if possible. (25f/(2f-10)) + (2f/(10f-50))

3) Divide and simplify. ((z^2-81)/z) / ((z-9)/(z+1))
%28%28%28z%5E2-81%29%2Fz%29%29+%2F+%28%28%28z-9%29%2F%28z%2B1%29%29%29 = %28%28z%5E2-81%29%2Fz%29+%2A%28%28z%2B1%29%2F+%28z-9%29%29 = %28z%5E2-81%29%28z%2B1%29%2F+%28z%28z-9%29%29=%28z-9%29%28z%2B9%29%28z%2B1%29%2F+%28z%28z-9%29%29 =%28%28z-9%29%2F%28z-9%29%29%28+%28z%2B9%29%28z%2B1%29%2Fz%29=+%28z%2B9%29%28z%2B1%29%2Fz
4) Find the LCM of z^2+12z+36 and z^2+3z-18. Use Factored Form. I think the answer is (z+3), is this correct? NO
+z%5E2%2B12z%2B36+=%28z%2B6%29%5E2
+z%5E2%2B3z-18=%28z%2B6%29%28z-3%29
The LCM is the least common multiple of those two expressions. In general, it must have all factors in the two expressions with the largest exponent seen in the two expressions.
The LCM must have z%2B6 and z-3 as factors to be a multiple of %28z%2B6%29%28z-3%29, but we need a 2 as the exponent of %28z%2B6%29 for the LCM to be a multiple of %28z%2B6%29%5E2.
So LCM=%28z%2B6%29%5E2%28z-3%29
5) Find the LCM of 21y^7 and 147y^9. I think this is 3y^7, is this right? NO
21y%5E7=3%2A7%2Ay%5E7
147y%5E9=3%2A49y%5E9=3%2A7%5E2%2Ay%5E9
A multiple of those two expressions must have 3, 7%5E2, and y%5E9 as factors, so
LCM=3%2A7%5E2%2Ay%5E9=147y%5E9
In fact, 147y%5E9=%2821y%5E7%29%2A7y%5E2%29+is+a+multiple+of+%7B%7B%7B21y%5E7
6) Simplify by removing by factors of 1. 42y^5/54y^3 Is this 7y^2/9? YES