Homework Help: math

Posted by ally on Saturday, January 21, 2012 at 6:46pm.

1)Which values of x are in the solution set
y=3x-1
y=x^2-5x+6

2)6ab (radical8b)-3a radical(32b^3)+9ab (radical5b)

3) write in simplest form (2/x+4/y)/(4x/y-y/x)

4)solve for all values of x
8x^3+4x^2-18x-9

5)abs(4x+6)-4x=-10

6) simplify 4x^-5y7 /(3x^3y^-8)^-3

math - yuuka, Sunday, January 22, 2012 at 12:12am
1) set the two equations equal to each other:
3x-1 = x^2-5x+6

Combine like terms:
x^2-8x+7 = 0

Factor:
(x-1)(x-7)=0

x = 1, 7
math - Steve, Sunday, January 22, 2012 at 11:24am
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/y-y/x)
= (2y+4x)/xy / (4x^2 - y^2)/xy
= (y+2x)/(2x-y)(2x+y)
= 1/(2x-y)

4)
8x^3 + 4x^2 - 18x - 9
= (2x+3)(2x-3)(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
-4x-6 - 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

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