how do you solve this:
2x-3 = (3+x)/2

- anonymous

how do you solve this:
2x-3 = (3+x)/2

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- anonymous

multiply by 2, subtract an x and add a 3

- anonymous

thank you

- anonymous

x=3

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## More answers

- anonymous

i need the process, not the answer which i already know

- anonymous

I don't think so

- anonymous

The point is, you want to get all your x terms on one side and your non-variable terms on the other

- anonymous

\[2*(2x-3)=[(3+x)/2]*2\]

- anonymous

then it goes to : 4x-6=3+x

- anonymous

thank you so much, i understand now

- anonymous

the get x by itself...so subtract -x to the left so next step: 3x=9

- anonymous

do i get a medal :)

- anonymous

yes!
but i don't know how to give one

- anonymous

i need help with this one too please:
2x-1= 1/3(5-3x) + 4

- anonymous

Ok, well why don't you tell us how to start this one

- anonymous

Oh, and is it
\[\frac{1}{3(5-3x)}\]
or
\[\frac{1}{3}(5-3x)\]

- anonymous

the second one

- anonymous

Ok. So which part do you want to tackle first? (There's lots of different paths to the same answer)

- anonymous

Overall goal is to get all the x terms on one side and the non-x terms on the other side

- anonymous

You tell me what to do, and I'll write the new equation.

- anonymous

\[2x-1= \frac{1}{3}(5-3x) + 4\]

- anonymous

i don't know how to start that's why i'm asking

- anonymous

That's ok. Just pick a term that is in the wrong place

- anonymous

We want everything with an x on the left, and everything without an x on the right

- anonymous

Pick a term, any term.

- anonymous

3x

- anonymous

Ah, that term is a little tricky because it's part of a factor in a product. We will need to expand that product out first before we can work with that term directly

- anonymous

But that's ok!

- anonymous

That just means we need to distribute the 1/3 to each of the terms in the other factor

- anonymous

So what do we get when we distribute that 1/3?

- anonymous

3-9x?

- anonymous

no. we have to multiply each term in the left factor by 1/3
\[\frac{1}{3}(5-3x) = \frac{1}{3}*5 - \frac{1}{3}*3x\]

- anonymous

Does that make sense? So what do we have when we simplify that product?

- anonymous

i'm not sure...

- anonymous

a(b+c) = ab + ac
Right? Basic multiplicative distribution

- anonymous

ok so would it be 5/3 -x?

- anonymous

yes

- anonymous

So now we have:
\[2x-1= \frac{5}{3}-x + 4\]

- anonymous

so the whole thing would be :
2x-1 = 5/3-x+4?

- anonymous

Yes

- anonymous

Can you solve from there? or do you need more help with it?

- anonymous

maybe a little more

- anonymous

Ok, so pick a term that's in the wrong place

- anonymous

how would i combine 5/3 + 4
that's the next step right?

- anonymous

You can certainly do that yes.

- anonymous

There is no 'next step'. There are a lot of ways to do it. If you want to do that next, go for it =)

- anonymous

ok so how do you do it?

- anonymous

How do you add \(\frac{5}{3} + 4\)

- anonymous

You have to change the denominator of the 4 to match the 3 in the 5/3

- anonymous

We do this using the trick of multiplying by 1.

- anonymous

\(4 \times 1 = ?\)

- anonymous

4

- anonymous

Ok, so if we multiply by 1 we won't have changed the value, right?

- anonymous

right

- anonymous

Ok, so what is \(\frac{3}{3}\)

- anonymous

1

- anonymous

So you're saying then that if I multiply 4 by 3/3 I will still have 4?

- anonymous

\[4\times \frac{3}{3} = \frac{4\times 3}{3} = \frac{12}{3}\]

- anonymous

And 12/3 does equal 4, but it's in a different form that is now easy to add to 5/3

- anonymous

Did you follow that?

- anonymous

yes
so the answer would be 17/3

- anonymous

Well, the sum of those two terms would be 17/3 yes

- anonymous

So we have:
\[2x-1= \frac{17}{3}-x\]

- anonymous

then irt would be 3x-1= 17/3

- anonymous

Certainly

- anonymous

it would be 3x = 1 17/3?

- anonymous

yeah, but I dun like mixed numbers

- anonymous

Can you do the same trick to the 1 we did to the 4 to make it easy to add to the 17/3

- anonymous

isn't it 20/9?

- anonymous

x = 20/9 is correct yes.

- anonymous

you don't divide that by 3?

- anonymous

Huh? we did already

- anonymous

It was 3x = 20/3

- anonymous

oh ok i thought it was 3x = 1 17/3? and then you divide by 3 on both sides to get x

- anonymous

\(3x = 1+\frac{17}{3}\)\[\implies 3x = 1\times \frac{3}{3} + \frac{17}{3}\]\[\implies 3x = \frac{1\times 3}{3} + \frac{17}{3}\]\[\implies 3x = \frac{3}{3} + \frac{17}{3}\]\[\implies 3x = \frac{20}{3}\]

- anonymous

Then you divide by 3 on both sides

- anonymous

Follow all that?

- anonymous

yes thanks

- anonymous

but how do you divide 3 on both sides, that what i want to know

- anonymous

dividing by 3 is the same as multiplying by 1/3

- anonymous

wouldn't the answer be different?

- anonymous

nope.
\[3x \times \frac{1}{3} = \frac{20}{3} \times \frac{1}{3}\]
\[x = \frac{20}{9} \]
\[3x \div 3 = \frac{20}{3} \div 3\]
\[x = 6.\bar{6}\bar{6} \div 3 = 2.\bar{2}\bar{2} = \frac{20}{9}\]

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