Posted by **jasmine20** on Tuesday, December 12, 2006 at 10:12pm.

can someone correct this for me or help me.

6x^2y^3+ 9x^2y^3 divided by 3x^2y^2

my solving and answer is:

15 x x y y y

---------- = 5y

3 x x y y

Looks good to me. I've never thought of showing x^2y^3 as xxyyy. That a neat wrinkle. We see quickly that the Xs cancel and all but one y in the top cancel.

so the way i did it is correct

as well as the answer

yes.

how do i solve 18-6(5x-8)=-24

I think I take -6X5X and-6X-8=

18-30x48=-24

18 18

-30x30=-24

div -30

=x 30=-24

-24

so x= 6

or am I all misses up

THIS IS WHAT YOU DO

18-6(5X-8)=-24

18-30X+48=-24 iN THIS STEP YOU DO DISTRUBITION SUCH AS -6(5X)=-30X, -6(-8)=48

SO

18-30X+48=-24

66-30X=-24 thIS STEP YOU ADD THE 18+48=66, AND YOU JUST BRING DOWN THE -30X

now you have 66-30x=-24

so then you bring the 66 to the other side of the equation which you have to do the opposite so it should look like this

-30x=-24-66

which will give you -30x=-90

NOw to get the x alone you divided bothe sides by -30

x=-90/-30

x=3

TO check you just substitute

18-6(5x-8)=-24

18-6(5(3)-8)=-24

18-6(15-8)=-24

18-6(7)=-24

18-42=-24

-24=-24

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