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Mathematics 9 Online
OpenStudy (babynini):

Can someone help me solve this? Please explain the process :)

OpenStudy (babynini):

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OpenStudy (fibonaccichick666):

So, what have you tried so far?

OpenStudy (babynini):

Well, I've gotten all the way to where I could factor it out, but it's coming out to weird numbers so I'm messing up somewhere in the process.

OpenStudy (fibonaccichick666):

ok, show me what you have done, and I can see what has occurred

OpenStudy (babynini):

Here, let me try and write out my process :) Just a minute. thanks!

OpenStudy (fibonaccichick666):

Don't be surprised if you have a fractional answer though

OpenStudy (fibonaccichick666):

try using the equation button or just typing instead of drawing. just use division and parentheses to show the fractions

OpenStudy (babynini):

\[4(y-1)(y+1)(5/y-1)-4(y-1)(y+1)(1/4)=4(y-1)(y+1)(3/y+1)\]

OpenStudy (babynini):

4(5)(y+1)-1(y-1)(y+1)=4(3)(y+1)

OpenStudy (babynini):

20y+20-y^2+1=12y+12

OpenStudy (fibonaccichick666):

ok so what was your first step?

OpenStudy (babynini):

finding the LCD

OpenStudy (fibonaccichick666):

ok, so my recommendation is to get rid of the fractions first(ie lcd), what was your lcd?

OpenStudy (babynini):

4(y-1)(y+1)

OpenStudy (fibonaccichick666):

good, now your multiplication of both sides by that lcd is where I see an error

OpenStudy (fibonaccichick666):

so I turned 1/4 into .25 so I didn't need the 4: \[\frac{5}{y-1}-.25=\frac{3}{y+1}\\ (y^2-1)[\frac{5}{y-1}-.25]=(y^2-1)[\frac{3}{y+1}]\\ 5(y+1)-.25(y^2-1)=3(y-1)\]

OpenStudy (fibonaccichick666):

so try it from there now. You just cancelled incorrectly, but you have the right idea. Let me know what you get

OpenStudy (fibonaccichick666):

you should get pretty whole number answers

OpenStudy (babynini):

Ah, sorry. I got confused with that.

OpenStudy (babynini):

You just didn't use .25 as part of the lcd?

OpenStudy (fibonaccichick666):

nope, it isn't necessary to. Since 1/4=.25, I didn't use it, but the issue was you had y+1 on the right hand side instead of y-1

OpenStudy (babynini):

Ah ok. Yeah, I actually did that just on here, but on my own work i used y-1 and the answer was still weird?

OpenStudy (babynini):

I got 0=y^2-8y-33

OpenStudy (fibonaccichick666):

that's right

OpenStudy (fibonaccichick666):

you just need to factor it now

OpenStudy (fibonaccichick666):

0=(y+___)(y-___)

OpenStudy (babynini):

oh! it is right? xD Dangit. I had that answer forever ago haha

OpenStudy (fibonaccichick666):

You just have to factor to get the 2 answers for y

OpenStudy (babynini):

(y-11)(y+3)

OpenStudy (babynini):

y= 11,-3

OpenStudy (fibonaccichick666):

That is math for ya, making you think you are wrong even when you are right,

OpenStudy (fibonaccichick666):

Yepp yepp

OpenStudy (babynini):

haha ikr. Thank you soo much. I'm sorry that took so long! You're a champ

OpenStudy (fibonaccichick666):

No problem, and thanks. Any time someone is willing to put the effort in, I have no problem helping. Feel free to tag me for assistance when needed :)

OpenStudy (babynini):

^-^

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