Question

Will Give Medal For Correct Answer

Please Show Work
Solve: x+12/x+4 = x/x+8

Answer 1

@skullpatrol can u help?

Answer 2

Can you think of any restrictions on this equation?

Answer 3

restrictions?

Answer 4

Yes, restrictions are limits on values that the variable can have.

Answer 5

\[\Large \frac{ x+12 }{ x+4 } = \frac{ x }{ x+8 }\]
yes...?

Answer 6

@skullpatrol yea, -4, 8, 6, -6
@agent0smith yup thats it

Answer 7

Why did you say -4?

Answer 8

the series of numbers are the restrictions

Answer 9

Can you give a reason why you chose those particular numbers?

Answer 10

they were pre chosen for me

Answer 11

@skullpatrol those are likely the answer choices.

@IanK16. he's asking what values of x will make the equation have a denominator of zero.

Answer 12

right and i need help with that

Answer 13

\[\Large \frac{ x+12 }{ x+4 } \] let's look at the left side first. What value of x will make the denominator (the bottom part) zero? ie... x+4 = 0

Answer 14

because... can you divide by zero?

Answer 15

-4 will make it zero

Answer 16

Correct. Now look at the right side of \[\Large \frac{ x+12 }{ x+4 } = \frac{ x }{ x+8 }\] what value of x will give a denominator of zero on the right?

Answer 17

a -8 will

Answer 18

Correct. So those are our "excluded" values, x=-4 and x=-8, in other words, x cannot be equal to those (they aren't solutions).

Now you have to solve it, keeping in mind if you get x=-4 or x=-8, they aren't valid solutions.

Answer 19

ok so what exactly would be the first step?

Answer 20

Cross multiply first... do you know how to do that?

\[\Large \frac{( x+12) }{( x+4) } = \frac{ x }{ (x+8) } \]

Answer 21

yes i do, im working on it right now

Answer 22

\[\frac{ 21x ^{2}+96 }{x ^{2}+4x}\]

Answer 23

@IanK16. What happened to the equals sign?

Answer 24

you should get \[\large (x+12)(x+8) = x( x+4)\]

Answer 25

when i cross multiply i should get that?

Answer 26

Yes, cross multiply like this

Answer 27

right and the answer above is what i got

Answer 28

Show us how you got \[\frac{ 21x ^{2}+96 }{x ^{2}+4x}\]

Answer 29

Note that in my picture, the equals sign stays where it is... http://assets.openstudy.com/updates/attachments/5170675ee4b0012aba7e2fea-agent0smith-1366325392859-crossmultiply.png

Answer 30

i cross multiplied.
(x+12)(x+8) then (x+4)(x)

Answer 31

Like I said @IanK16. What happened to the equals sign?

Answer 32

You started with an equation, and now you only have an expression.

Answer 33

idk, would it still = x/(x+8)

Answer 34

http://www.mathsisfun.com/algebra/images/cross-multiply.gif

Answer 35

alright, but i apparently did my math wrong somewhere cuz @agent0smith got (x+12)(x+8) = x(x+4)

Answer 36

it looks more like u factored it

Answer 37

I didnt factor it... i just didn't expand it/multiply it out. Do it exactly as shown in this pic: http://www.mathsisfun.com/algebra/images/cross-multiply.gif

Answer 38

And show us what you get.

Answer 39

\[\Large \frac{( x+12) }{( x+4) } = \frac{ x }{ (x+8) }\]

Answer 40

\[21x ^{2}+96 = x ^{2}+4x\]

Answer 41

Work out the LHS: (x + 12)(x +8) using FOIL.

Answer 42

yea i did thats what i got

Answer 43

How are you getting that left hand side? the right hand side is correct.

I have no idea how you got the left, how are you getting 21x^2?.... immediately after cross multiplying, you should have \[\large (x+12)(x+8) = x( x+4)\]

Answer 44

Where does the 21 come from?

Answer 45

when you multiply thos out you get x^2+8x+12x+96, then i added like terms

Answer 46

That's correct.. but where did the 21 come from...?

Answer 47

\[\large x^2+8x+12x+96 \] note that x^2 and x are *not* like terms. You can't add an x^2 term to an x term.

Answer 48

8x + 12x = 20x then 20x + x^2 is 21x^2

Answer 49

20x + x^2 is 20x + x^2. Just like 2+x is 2+x, you can't combine unlike terms.

Answer 50

alright so what your saying is that it should say: x^2 + 20x + 96 ?

Answer 51

correct

Answer 52

ok i apparently forgot how do this, can u walk me through the next step?

Answer 53

Yes, you need to make sure you don't ever combine x^2 with x. You can't combine terms unless they're exactly the same, like you can't combine x^3 with x^2 for example.

Answer 54

\[\large x^2+20x+96 = x^2 +4x\]

Answer 55

ok so now i put the constants one side?

Answer 56

so then i would get 16x = 96?

Answer 57

Do it one step at a time.

First, subtract x^2 from both sides

\[\large x^2+20x+96 = x^2 +4x \]
\[\large 20x+96 = 4x\]
now subtract 4x from both sides.

Answer 58

that would be -4x+20x+96=0

Answer 59

Correct. Now you can combine like terms, and then subtract 96 from both sides.

Answer 60

it would be 16x=-96 then

Answer 61

Correct.

Answer 62

then i divide to get -6?

Answer 63

Yep

Answer 64

alright thanks you've been a huge help!

Answer 65

No prob :)