5+3[1+2(2x-3]=6(x+5)

- anonymous

5+3[1+2(2x-3]=6(x+5)

- jamiebookeater

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- sandra

I'm assuming you mean 5+3(1+2(2x-3)) = 6(x+5)

- sandra

this is another case where you need to distribute

- sandra

start with the right side, what is 6(x+5) ?

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## More answers

- anonymous

yes

- anonymous

6x+30?

- sandra

yep!

- sandra

ok now start from the inside out on the left hand side

- anonymous

do u always start with the right side

- sandra

so what is 2(2x-3)

- sandra

well, I think generally start from inside out

- sandra

as in inner-most thing that needs to be expanded

- anonymous

4x-6

- sandra

ok great, so to recap, so far we have:
5 + 3(1 + 4x - 6) = 6x + 30

- anonymous

gotcha

- sandra

so to make it easier on yourself, how about combing like terms in the parens, so what is (1+4x - 6) ?

- anonymous

5x-6 or do u break that down further

- sandra

well, that's not correct

- sandra

you can't combine 1 and 4x, since you don't know what 4x

- sandra

but you can combine 1 and -6

- sandra

(by adding them)

- anonymous

-5

- sandra

if it had been (1x +4x -6) you would have been correct, since the 1 and 4 are both x's

- sandra

right so the middle is now -5 + 4x, or 4x - 5

- sandra

so now you have : 5 + 3(4x - 5) = 6x + 30

- sandra

what is your next step?

- anonymous

5+12x-15=6x+30

- sandra

great =)

- sandra

so can you combine anything on the left side?

- anonymous

5 and 15 which is -10

- sandra

yep =), so now what do you have?

- anonymous

is it 12x-10=6x+30

- sandra

exactly

- sandra

so now the basic rule of algebra, and equations/equalities in general - you can perform any operation to both sides of an equation and still have the equatoin be true (satisfied)

- sandra

the reason that's powerful is because it allows you to eventually get x all by itself

- sandra

so let's say you wanted to get 12x by itself on the left hand side

- sandra

what could you add to both sides to make that happen?

- anonymous

add 10

- sandra

yep, and what do you get if you add 10 to both sides

- anonymous

12x=6x+40

- sandra

exactly. ok , so remember, when you're solving for a variable (in this case x), the idea is to get all of that variable to one side, and then eventually just one instance of that variable by itself (x)

- sandra

so what could you subtract from both sides, to get all of the x's in one place?

- anonymous

would u minus 6x

- sandra

exactly, and then what you get?

- anonymous

6x=40 then i divided by 6 both sides

- sandra

yep!

- sandra

good work

- anonymous

6.6?

- sandra

yeah, or you could write it as a fraction

- sandra

so x = 40/6

- sandra

and that can be reduced if you divide both top and bottom by 2

- sandra

so x = 20/3

- sandra

so now that you think x = 20/3

- anonymous

thank u so much i think im getting it sum what lol

- sandra

you stick it back into the original equation (or just use 6.666666 to check in your calculator)

- sandra

and if the equation is true, then you know you got it right

- sandra

so for these types of problems, try expanding everything out first

- sandra

and then just keep doing the same operations on both sides until you conveniently get all the x's (or whatever you're solving for) in one place

- sandra

and then just get the value for one x by dividing

- sandra

after a while, you'll start seeing some shortcuts if they're there

- sandra

for example, if you had 3(x+3) = 12

- sandra

you wouldn't *have* to expand first

- sandra

you could just divide both sides by 3

- sandra

to get x+3 = 4

- sandra

and then subtract 3 from both sides to get x=1

- sandra

but seeing that will come with practice

- sandra

good luck!

- anonymous

oh ok thanks a bunch !!!!

- sandra

yep np

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