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first use the distributive property twice
\[a(b+c)=ab+ac\]
tell me what u get?

*thrice

9x+10x+55 = 3x+63 + 16x-6

CLT.. combine like terms, then solve for x now....

correct,now combine like terms
9x,10x,3x,16x
55,63,-6

Then you get: 19x+55 = 19x+57

Would this be a no solution? Since 19x-19x = 0?

thats correct,so now that 19x gets cancelled
and there will be no solution

nice work...

So the answer is "No Solution" ?

yup :)

Okayy. I need help with 2 more problems.

sure,go ahead and ask

what u got?

I don't know how to solve this..

1.divide by W
2.subtract 2
3.divide by z
all these to both sides

You don't distribute W first?

that would complicate it,though possible.

This is what I got:
G = W (2+zp)
G/W = W(2+zp)/W
G/W – 2 = zp
G/W/z – 2 = p

ok,in last step,u missed dividing z to 2
also
G/W/z=G/Wz

So it would be: G/Wz - 2/z = p ?

yup,which u can write as
\[\frac{G-2W}{Wz}\]

Okay...one more problem, please.