Question

proportion- solve x-1/x+1=2/3x... i know you cross multiply to get (x-1)(3x)=(x+1)(2)...but then get stuck

Answer 1

That's a great start.

Use the distributive property to start simplifying each side of the equation you have.

a(b + c) = ab + ac

I'll do the left side of the equation for you.

(3x)(x - 1) = (3x)(x) + (3x)(-1) = 3x^2 - 3x

Now do the same for the right hand side and then solve for x.

Answer 2

i'm confused..

Answer 3

dont you cross multiply then factor out or something?

Answer 4

No factoring. What mkuehn has shown you is the left side distributed. The same needs to be done on the right.

Answer 5

So the right would be:
\[2(x+1)=2x+2\]

Answer 6

Then you bring evrything to one side to solve for 0.

Answer 7

So your setup for the final solution would be:
\[3x ^{2}-3x=2x+2\]

Answer 8

i somehow got a squared x... i did 3x(x-1)=2(x+1) which brought me to 3x^2-3=2x+1

Answer 9

or did i make that up

Answer 10

You almost have it.

Answer 11

3x * x = 3x^2
3x * -1 = ?

Answer 12

so what i have now is 3x^2-3x=2x+2.. had some errors on my part

Answer 13

Okay, so now bring the terms from the right hand side to the left hand side by subtracting them.

Answer 14

then i add 3x to both sides?

Answer 15

Yes, that's where I dropped off earlier. When doing on paper. I like to draw a line from the itkem being distributed to each item that needs to be multiplied so I can keep it straight.

Answer 16

You can move everything to whichever side makes it easier on you, but they all need to be on one side.

Answer 17

You can add 3x to both sides, but I think it is easier to work with the term that is squared as a positive, so bring everything to the left hand side.

Answer 18

So if you add 3x to each side you also need to subtract 3x^2 as well.

Answer 19

I agree with mkuehn on that. I prefer positive x^2 numbers as well :)

Answer 20

Once you have them all on the same side you can factor if possible.

Answer 21

Looking at this one I don't see an easy factoring so you're probably going to have to apply the quadratic formula.

Answer 22

3x^2-5x-2=0?

Answer 23

That's what I have as well.
So ax^2+bx+c=0
Have you worked with the quadratic formula before?

Answer 24

long time ago

Answer 25

So we have:
\[(-b \pm \sqrt{b ^{2}-4ac})/2a\]

Answer 26

So we look at our current work. Can you find a,b,c?

Answer 27

b=3 a=5 c=-2?

Answer 28

Not quite :)
The form is:
\[ax ^{2}+bx+c=0\]

Answer 29

ax^2+bx+c=0

3x^2-5x-2=0

Answer 30

Go for it again.

Answer 31

Keep in mind the signs matter. So a -5 will be a -5 in the quadratic.

Answer 32

You're about 2 steps from your answers :)

Answer 33

okay, do i add the -5x and -2?

Answer 34

You need to get the values of a,b and c and then plug them into teh quadratic equation listed above. I'm trying to step you through getting a,b and c first. Then we'll plug them in.

Answer 35

See how neatly they line up on mkuehn's last post. He's made it easier for you.

Answer 36

If you give up I can give you a,b and c. I'm not trying to torture you. Just trying to heklp you through the process.

Answer 37

yeah i understand that im just getting more confused as we go on....i'm trying to hard =[

Answer 38

Okay looking at mkuehns post,
a = 3
b = -5
c= -2

Answer 39

Can you see that looking at it again?

Answer 40

You've honestly done the hardest part earlier putting everything all on the same side :)

Answer 41

Lots of folks get lost of the cross multiplying.

Answer 42

So now we can go back to my earlier part and plug the numbers in:
\[(-(-5)\pm \sqrt{(-5)^{2}-4(3)(-2)})/2(3)\]

Answer 43

so -5+or- -5^2-4(3*-2)/2{3}?

Answer 44

Simplifying we get:
\[(5\pm \sqrt{25+24})/6\]

Answer 45

Yes. Since we have x^2 in the problem there will be two answers. Thus the plus or minus part.

Answer 46

Note how the -(-5) became 5. And the (-5)^2 became 25.

Answer 47

That's why I mentioned the signs mattering.

Answer 48

o ok

Answer 49

So one more step and we'll have the two final answers:
\[(5\pm \sqrt{49})/6\]
So our answers are:
\[(5+\sqrt{29})/6\]
And
\[(5-\sqrt{49})/6\]

Since \[\sqrt{49}=7\]
we can simplify some.

Answer 50

That root 29 is supposed to be 49
I typoed

Answer 51

\[(5+7)/6 = 12/6 = 2\]

Answer 52

\[(5-7)/6 = -2/6 = -1/3\]

Answer 53

So clear as mud? :P As you can see there was no way we could have factored out those answers.

Answer 54

i dont get the last 2 posts

Answer 55

\[\sqrt{49}\]

Answer 56

=7

Answer 57

So if you step up a set and ignore my 29 typo we replaced root 49 with 7 for both of our answers.

Answer 58

The easy way to test is to plug the answers back into:
\[3x ^{2}-5x-2=0\]
and make sure you get 0.
So the first answer was 2:
\[3(2)^{2}-5(2)-2=3(4)-10-2=12-12=0\]

Answer 59

The second answer was -1/3:
\[3(-1/3)^{2}-5(-1/3)-2=3(1/9)+5/3-2=3/9+5/3-2=1/3+5/3-2=6/3-2=2-2=0\]

Answer 60

That last part that got cut off is:
\[6/3-2=2-2=0\]

Answer 61

i dont even know what i'm looking at

Answer 62

Okay we solved using the quadratic equation with a, b, and c that you helped find.

Answer 63

We plugged them into that formula with the plus and minus which we separated into two parts.

Answer 64

You're looking at magi... err math. :)

Answer 65

When we solved those parts we found the answers to be 2 and -1/3.

Answer 66

From there we wanted to make sure we found the right answers and didn't make a mistake so we plugged it back into the original formula.

Answer 67

using the formula

Answer 68

That what the last part was. I know it all looks intimindating, but so does a multicourse meal. Small bites and you'll get there :)

Answer 69

I have to roll. I know it seems a little intimidating but it's not as bad as it looks. Doing a few more of the problems may make it make more sense. When I come across math problems that leave me lost I head over to khanacademy.org. You may want to give it a look see. Sometimes things make more sense when you see them visually solved.

Answer 70

yeah i'll try that. thank you very much though for spending so much time on that problem