Question 576908
1)	Subtract. Simplify if possible. (17cb/(c^2-b^2)) - ((c-b)/(c+b)) 
{{{17cb/(c^2-b^2)- (c-b)/(c+b)}}}={{{17cb/((c - b)(c+b))- (c - b)/(c+b)}}}={{{17cb/((c - b)(c+b))}}} - {{{(c - b)(c-b)/((c+b)(c-b))}}}={{{(17cb-(c-b)(c-b))/((c+b)(c-b))}}}={{{(17cb-(c^2-2bc+b^2)) /((c+b)(c-b))}}}={{{(19cb-c^2-b^2)/((c+b)(c-b))}}}
2) Add. Simplify if possible. (25f/(2f-10)) + (2f/(10f-50)) 
{{{25f/(2f-10)+2f/(10f-50)=5*25f/(5*(2f-10))+2f/(10f-50)= (125f+2f)/(10f-50)=127f/(10f-50)=127f/10/(f-5)}}}
3) Divide and simplify. ((z^2-81)/z) / ((z-9)/(z+1)) 
{{{(((z^2-81)/z)) / (((z-9)/(z+1)))}}} = {{{((z^2-81)/z) *((z+1)/ (z-9))}}} = {{{(z^2-81)(z+1)/ (z(z-9))}}}={{{(z-9)(z+9)(z+1)/ (z(z-9))}}} ={{{((z-9)/(z-9))( (z+9)(z+1)/z)= (z+9)(z+1)/z}}}
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^2+12z+36 =(z+6)^2}}}
{{{ z^2+3z-18=(z+6)(z-3)}}}
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+6}}} and {{{z-3}}} as factors to be a multiple of {{{(z+6)(z-3)}}}, but we need a 2 as the exponent of {{{(z+6)}}} for the LCM to be a multiple of {{{(z+6)^2}}}.
So LCM={{{(z+6)^2(z-3)}}}
5) Find the LCM of 21y^7 and 147y^9. I think this is 3y^7, is this right?    NO
{{{21y^7=3*7*y^7}}}
{{{147y^9=3*49y^9=3*7^2*y^9}}}
A multiple of those two expressions must have {{{3}}}, {{{7^2}}}, and {{{y^9}}} as factors, so
LCM={{{3*7^2*y^9=147y^9}}}
In fact, {{{147y^9=(21y^7)*7y^2) is a multiple of {{{21y^7}}}
6) Simplify by removing by factors of 1. 42y^5/54y^3 Is this 7y^2/9?    YES
{{{42y^5/54y^3=6*7*y^3*y^2/(6*9*y^3)=(6/6)( y^3/y^3)(7y^2/9)= 7y^2/9}}}