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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