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anonymous
 one year ago
I cant seem to figure this out help please :)
What is the simplified form of 2 over x squared plus x minus 1 over x ?
anonymous
 one year ago
I cant seem to figure this out help please :) What is the simplified form of 2 over x squared plus x minus 1 over x ?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2 }{ x^2+x }\frac{ 1 }{ x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 @MTALHAHASSAN2 @markaskingalexandria1 @animalfan1 @abs202 @Summersnow8 @RedNeckOutLaw

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439243455162:dw

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439243547714:dw

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439243586590:dw

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439243628166:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1The first step is to factor the left denominator. Can you do that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's not one of the answers want me too post them?

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 is my answer right

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Once again, first, factor the left denominator.

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0how is it wrong?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x minus 1 over the quantity x times x plus 1 1 minus x over the quantity x times x plus 1 3 minus x over the quantity x times x minus 1 x plus 2 over the quantity x times x minus 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so \[\frac{ 2 }{ x(x+1)}\]?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1This problem is a subtraction of fractions. We need a common denominator. The way to find a common denominator is to first factor all denomiators. \(\Large \dfrac{ 2 }{ x^2+x }\dfrac{ 1 }{ x }\) Factor the left denominator. Correct. \(\Large =\dfrac{ 2 }{ x(x+1) }\dfrac{ 1 }{ x }\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Now you need to find the LCD of x(x + 1) and x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so x(x+1) is the lcd so multiply 1/x by x+1?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Correct. Multiply 1/x by (x + 1)/(x + 1) \(\Large =\dfrac{ 2 }{ x(x+1) }\dfrac{ 1 }{ x } \times \dfrac{x + 1}{x + 1} \)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large =\dfrac{ 2 }{ x(x+1) }\dfrac{x + 1}{x(x + 1)} \)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Now do the actual subtraction. Write a new fraction with the common denominator, and subtract the numerators. Remember to enclose the second numerator in parentheses.

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 huh i goted

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0so first you have to factor that x square

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large =\dfrac{ 2  (x + 1)}{ x(x+1) } \) \(\Large =\dfrac{ 2  x  1}{ x(x+1) } \) \(\Large =\dfrac{ 1  x}{ x(x+1) } \)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Numerator is either 1  x or x + 1, not x  1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sweet! That's one of the answers! THANK YOU I GET IT :)!!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1The first step was factoring x^2 + x

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Great. You're welcome.

MTALHAHASSAN2
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 nice work!!
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