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I'm assuming you mean 5+3(1+2(2x-3)) = 6(x+5)
this is another case where you need to distribute
start with the right side, what is 6(x+5) ?

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Other answers:

ok now start from the inside out on the left hand side
do u always start with the right side
so what is 2(2x-3)
well, I think generally start from inside out
as in inner-most thing that needs to be expanded
ok great, so to recap, so far we have: 5 + 3(1 + 4x - 6) = 6x + 30
so to make it easier on yourself, how about combing like terms in the parens, so what is (1+4x - 6) ?
5x-6 or do u break that down further
well, that's not correct
you can't combine 1 and 4x, since you don't know what 4x
but you can combine 1 and -6
(by adding them)
if it had been (1x +4x -6) you would have been correct, since the 1 and 4 are both x's
right so the middle is now -5 + 4x, or 4x - 5
so now you have : 5 + 3(4x - 5) = 6x + 30
what is your next step?
great =)
so can you combine anything on the left side?
5 and 15 which is -10
yep =), so now what do you have?
is it 12x-10=6x+30
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)
the reason that's powerful is because it allows you to eventually get x all by itself
so let's say you wanted to get 12x by itself on the left hand side
what could you add to both sides to make that happen?
add 10
yep, and what do you get if you add 10 to both sides
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)
so what could you subtract from both sides, to get all of the x's in one place?
would u minus 6x
exactly, and then what you get?
6x=40 then i divided by 6 both sides
good work
yeah, or you could write it as a fraction
so x = 40/6
and that can be reduced if you divide both top and bottom by 2
so x = 20/3
so now that you think x = 20/3
thank u so much i think im getting it sum what lol
you stick it back into the original equation (or just use 6.666666 to check in your calculator)
and if the equation is true, then you know you got it right
so for these types of problems, try expanding everything out first
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
and then just get the value for one x by dividing
after a while, you'll start seeing some shortcuts if they're there
for example, if you had 3(x+3) = 12
you wouldn't *have* to expand first
you could just divide both sides by 3
to get x+3 = 4
and then subtract 3 from both sides to get x=1
but seeing that will come with practice
good luck!
oh ok thanks a bunch !!!!
yep np

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