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−53>−3(3z+3)+3z
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You need to distribute on the right side.
What is the correct way to distribute
\(a(bx+c)=(ab)x+ac\) \(-3(3z+3)=(-3\times3)x+(-3\times3)\)
What would be the second step?
Post what you have as an answer for the distribution. Then combine like terms :)
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I'm really struggling with the distribution because there's both positive and negative numbers
Two things. 1) Not the place for that 2) You're*
−53>−3(3z+3)+3z
Ok, lets break it down. \(-3\times 3z=~?\) \(-3\times3=~?\)
-9z and -9
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Correct so now we have, \(-53>-9z+3z-9\) When you combine like terms, what do you get?
I got it. It is 22/3.
Thanks.
Correct!
Thanks.
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You are very welcome!
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