# 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

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> 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      Log On

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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(1864)   (Show Source): You can put this solution on YOUR website!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? NO 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 and as factors to be a multiple of , but we need a 2 as the exponent of for the LCM to be a multiple of . So LCM= 5) Find the LCM of 21y^7 and 147y^9. I think this is 3y^7, is this right? NO A multiple of those two expressions must have , , and as factors, so LCM= In fact, 6) Simplify by removing by factors of 1. 42y^5/54y^3 Is this 7y^2/9? YES