Expert: Richard J. Raridon - 4/6/2009

Question
Hi,
Can you help me with the following.
This test is due today and I only have these 4 left to finish. Sorry for the late notice.

All the problems are set up as a line being under the top of each equation, showing that it is a fraction or division of each of the numbers underneath each. I hope this makes sense.

#1
Multiply

16 – x^2    * x^2 +5x + 6
6x + 12       x^2 -8x + 16





#2
Simplify

3x^2 + 2  -  9x – x^2
x^2 – 4      x^2 - 4






#3
Solve

3 =    18   
      3x – 4


#4
Solve

 x     + x   = 3x – 4
x – 3            x – 3


Answer
#1You could have written it[(16-x^2)/(6x+12)][(x^2+5x+6)/(x^2-8x+16)]
so then you have [(4-x)(4+x)/6(x+2)][(x+2)(x+3)/(x-4)^2]
after cancelling you have [(4+x)(x+3)/{6(4-x)}]
#2 You should have written it (3x^2+2)/(x^2-4) -(9x-x^2)/(x^2-4)
which gives you (4x^2-9x+2)/(x^2-4) = (4x-1)(x-2)/[(x-2)(x+2)]
= (4x-1)/(x+2)
I can't figure out what you mean by #3 and #4.
Please re-write them.