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Mathematics 14 Online
OpenStudy (anonymous):

1. x+1 2 ---- - ----- x^2+5x-14 3x+21 2. x^2-49 x 2x-x^2 ------- ------- x^2-9x+14 9x^3-3x^4

OpenStudy (mathmale):

In the first case, you're subtracting two fractions, which happen to have different denominators. The very first thing to do here is to factor both given denominators and then to determine the least common denominator (LCD). In the secon case, you're multiplying two fractions. Again, I'd suggest you start by factoring everything. Then cancel out any like factors. Then multiply.

OpenStudy (anonymous):

@mathmale for the second one i got (x-7) (x+7) 2x(x) ---------- x ------ (x-7) (x+2) 3x^3 (3-x) but theres only one thing to factor

OpenStudy (anonymous):

@DSS can you help? sorry for all the questions!

OpenStudy (anonymous):

It's OK

OpenStudy (anonymous):

thanks!! i had like 80 questions and there were a few i didnt get

OpenStudy (anonymous):

x+1 2 ---- - ----- x^2+5x-14 3x+21 denominators: (x+7)(x-2) 3(x+7) Take care of the numerators now since the denominator will be 3(x+7)(x-2)

OpenStudy (mathmale):

@xokatexo : regarding the 2nd problem: I agree with your factoring, except in the case where you typed 2x(x). Mind going back and checking that?

OpenStudy (anonymous):

3(x+1) + -2((x-2) = 3x+3-2x+4 = x+7

OpenStudy (anonymous):

|dw:1388873629541:dw|

OpenStudy (anonymous):

how did that factor to be 1?

OpenStudy (anonymous):

OpenStudy (anonymous):

Are you asking me?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

go up to my picture 5 above

OpenStudy (anonymous):

thats just the original problem though

OpenStudy (anonymous):

3(x+1)+-2(x-2) = 3x+3-2x+4 = x+7

OpenStudy (anonymous):

ohh wait no i was talking about the 2nd problem

OpenStudy (anonymous):

x+7 is the simplified numerator. I put it over the denominator of 3(x+7)(x-2)

OpenStudy (anonymous):

i understand the first problem thank you you were very helpful

OpenStudy (anonymous):

but i dont understand the second one

OpenStudy (anonymous):

The x+7 cancels out in the numerator and denominator. Leaving 1 in the numerator

OpenStudy (anonymous):

k

OpenStudy (anonymous):

You have to factor lots. first fraction (x+7)(x-7)/(x-7)(x+2)

OpenStudy (anonymous):

yeah i got that

OpenStudy (anonymous):

second fraction x(2-x)/3x^3(3-x)

OpenStudy (anonymous):

|dw:1388874156329:dw|

OpenStudy (anonymous):

Now combine and multiply

OpenStudy (anonymous):

Now multiply. Computer is dying...I'll be off for a minute

OpenStudy (anonymous):

numerator (x-7)(2-x) denominator (x+2)(3x^2)(3-x) doesn't simplify any further

OpenStudy (anonymous):

sorry it took so long

OpenStudy (anonymous):

thank you so much!! sorry it wouldnt let me send that before

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