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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