Posted by ally on .
1)Which values of x are in the solution set
y=3x1
y=x^25x+6
2)6ab (radical8b)3a radical(32b^3)+9ab (radical5b)
3) write in simplest form (2/x+4/y)/(4x/yy/x)
4)solve for all values of x
8x^3+4x^218x9
5)abs(4x+6)4x=10
6) simplify 4x^5y7 /(3x^3y^8)^3

math 
yuuka,
1) set the two equations equal to each other:
3x1 = x^25x+6
Combine like terms:
x^28x+7 = 0
Factor:
(x1)(x7)=0
x = 1, 7 
math 
Steve,
2)
6ab√8b  3a√32b^3 + 9ab√5b
= 6ab*2√2b  3a(4b√2b) + 9ab√5b
= 12ab√2b  12ab√2b + 9ab√5b
= 9ab√5b
3)
(2/x+4/y)/(4x/yy/x)
= (2y+4x)/xy / (4x^2  y^2)/xy
= (y+2x)/(2xy)(2x+y)
= 1/(2xy)
4)
8x^3 + 4x^2  18x  9
= (2x+3)(2x3)(2x+1)
x = 3/2, 1/2, 3/2
5)
4x+6  4x = 10
If 4x+6 >= 0, 4x+6 = 4x+6 and you have
4x+6  4x = 10
6 = 10
no solutions there
If 4x+6 < 0, 4x+6 = (4x+6) and you have
4x6  4x = 10
8x = 16
x = 2
But, if x = 2, 4x+6 > 0, so no solution there
no solutions there
6)
4x^5 y^7 /(3x^3 y^8)^3
4x^5 y^7 * (3x^3 y^8)^3
4x^5 y^7 * 27x^9 y^24
108x^4 y^17
or, 108x^4/y^17