can you
3y(x-3)-2(x-3)
Combine 3y[2(x-3)]-2
3y[2x-6]-2 - um pretty sure its simplified...
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i might be wrong on this part so dont be reliable on this part.
3y(2x)*3y(12)= Yeah um dont ask me im in only in Algebra 1 Honors
You are in fact one step away from one important step.
This is already factored into groups nicely.
So this is what you need to do:
Put BOTH GCF ( 3y and -2) of the (x-3) as one factor. Like this:
(3y-2)(x-3).
This process is called factoring by grouping.
ok, it's already partially factored for you. You have two items multiplied by (x-3) so that's one factor. The other factor is 3y-2.
The final answer is (3y-2)(x-3)
Distribute 3(x-2) and 5(2x+3), giving you 3x-6+10x+15=35.
3y(x – 3) – 2(x – 3)
3y+3x-9y-2x+6
-6y+x
i think... been awhile since doing this. hope it helps.. and i hope im right. idk? lol
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3y(x-3)-2(x-3)
Combine 3y[2(x-3)]-2
3y[2x-6]-2 - um pretty sure its simplified...
______________________________________________
i might be wrong on this part so dont be reliable on this part.
3y(2x)*3y(12)= Yeah um dont ask me im in only in Algebra 1 Honors
You are in fact one step away from one important step.
This is already factored into groups nicely.
So this is what you need to do:
Put BOTH GCF ( 3y and -2) of the (x-3) as one factor. Like this:
(3y-2)(x-3).
This process is called factoring by grouping.
ok, it's already partially factored for you. You have two items multiplied by (x-3) so that's one factor. The other factor is 3y-2.
The final answer is (3y-2)(x-3)
Distribute 3(x-2) and 5(2x+3), giving you 3x-6+10x+15=35.
3y(x – 3) – 2(x – 3)
3y+3x-9y-2x+6
-6y+x
i think... been awhile since doing this. hope it helps.. and i hope im right. idk? lol