anonymous
  • anonymous
how do you solve this: 2x-3 = (3+x)/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
multiply by 2, subtract an x and add a 3
anonymous
  • anonymous
thank you
anonymous
  • anonymous
x=3

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anonymous
  • anonymous
i need the process, not the answer which i already know
anonymous
  • anonymous
I don't think so
anonymous
  • anonymous
The point is, you want to get all your x terms on one side and your non-variable terms on the other
anonymous
  • anonymous
\[2*(2x-3)=[(3+x)/2]*2\]
anonymous
  • anonymous
then it goes to : 4x-6=3+x
anonymous
  • anonymous
thank you so much, i understand now
anonymous
  • anonymous
the get x by itself...so subtract -x to the left so next step: 3x=9
anonymous
  • anonymous
do i get a medal :)
anonymous
  • anonymous
yes! but i don't know how to give one
anonymous
  • anonymous
i need help with this one too please: 2x-1= 1/3(5-3x) + 4
anonymous
  • anonymous
Ok, well why don't you tell us how to start this one
anonymous
  • anonymous
Oh, and is it \[\frac{1}{3(5-3x)}\] or \[\frac{1}{3}(5-3x)\]
anonymous
  • anonymous
the second one
anonymous
  • anonymous
Ok. So which part do you want to tackle first? (There's lots of different paths to the same answer)
anonymous
  • anonymous
Overall goal is to get all the x terms on one side and the non-x terms on the other side
anonymous
  • anonymous
You tell me what to do, and I'll write the new equation.
anonymous
  • anonymous
\[2x-1= \frac{1}{3}(5-3x) + 4\]
anonymous
  • anonymous
i don't know how to start that's why i'm asking
anonymous
  • anonymous
That's ok. Just pick a term that is in the wrong place
anonymous
  • anonymous
We want everything with an x on the left, and everything without an x on the right
anonymous
  • anonymous
Pick a term, any term.
anonymous
  • anonymous
3x
anonymous
  • 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
  • anonymous
But that's ok!
anonymous
  • anonymous
That just means we need to distribute the 1/3 to each of the terms in the other factor
anonymous
  • anonymous
So what do we get when we distribute that 1/3?
anonymous
  • anonymous
3-9x?
anonymous
  • 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
  • anonymous
Does that make sense? So what do we have when we simplify that product?
anonymous
  • anonymous
i'm not sure...
anonymous
  • anonymous
a(b+c) = ab + ac Right? Basic multiplicative distribution
anonymous
  • anonymous
ok so would it be 5/3 -x?
anonymous
  • anonymous
yes
anonymous
  • anonymous
So now we have: \[2x-1= \frac{5}{3}-x + 4\]
anonymous
  • anonymous
so the whole thing would be : 2x-1 = 5/3-x+4?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Can you solve from there? or do you need more help with it?
anonymous
  • anonymous
maybe a little more
anonymous
  • anonymous
Ok, so pick a term that's in the wrong place
anonymous
  • anonymous
how would i combine 5/3 + 4 that's the next step right?
anonymous
  • anonymous
You can certainly do that yes.
anonymous
  • 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
  • anonymous
ok so how do you do it?
anonymous
  • anonymous
How do you add \(\frac{5}{3} + 4\)
anonymous
  • anonymous
You have to change the denominator of the 4 to match the 3 in the 5/3
anonymous
  • anonymous
We do this using the trick of multiplying by 1.
anonymous
  • anonymous
\(4 \times 1 = ?\)
anonymous
  • anonymous
4
anonymous
  • anonymous
Ok, so if we multiply by 1 we won't have changed the value, right?
anonymous
  • anonymous
right
anonymous
  • anonymous
Ok, so what is \(\frac{3}{3}\)
anonymous
  • anonymous
1
anonymous
  • anonymous
So you're saying then that if I multiply 4 by 3/3 I will still have 4?
anonymous
  • anonymous
\[4\times \frac{3}{3} = \frac{4\times 3}{3} = \frac{12}{3}\]
anonymous
  • anonymous
And 12/3 does equal 4, but it's in a different form that is now easy to add to 5/3
anonymous
  • anonymous
Did you follow that?
anonymous
  • anonymous
yes so the answer would be 17/3
anonymous
  • anonymous
Well, the sum of those two terms would be 17/3 yes
anonymous
  • anonymous
So we have: \[2x-1= \frac{17}{3}-x\]
anonymous
  • anonymous
then irt would be 3x-1= 17/3
anonymous
  • anonymous
Certainly
anonymous
  • anonymous
it would be 3x = 1 17/3?
anonymous
  • anonymous
yeah, but I dun like mixed numbers
anonymous
  • 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
  • anonymous
isn't it 20/9?
anonymous
  • anonymous
x = 20/9 is correct yes.
anonymous
  • anonymous
you don't divide that by 3?
anonymous
  • anonymous
Huh? we did already
anonymous
  • anonymous
It was 3x = 20/3
anonymous
  • anonymous
oh ok i thought it was 3x = 1 17/3? and then you divide by 3 on both sides to get x
anonymous
  • 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
  • anonymous
Then you divide by 3 on both sides
anonymous
  • anonymous
Follow all that?
anonymous
  • anonymous
yes thanks
anonymous
  • anonymous
but how do you divide 3 on both sides, that what i want to know
anonymous
  • anonymous
dividing by 3 is the same as multiplying by 1/3
anonymous
  • anonymous
wouldn't the answer be different?
anonymous
  • 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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