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Mathematics 6 Online
OpenStudy (anonymous):

Can someone please solve? I really appreciate it! I know the answer, just dont know how to get to there!

OpenStudy (anonymous):

OpenStudy (lgbasallote):

are those choices below? or are they conditions?

OpenStudy (anonymous):

we identify the lcd first, then, we multiply it to both sides.

OpenStudy (anonymous):

By the way, we can still factor 2x-2. r

OpenStudy (anonymous):

(x-1)(x+2)

OpenStudy (anonymous):

first take lcm and cross multiply, [ 2x(x+2)+3(x-1) ](2x-2)=-1(x-1)(x+2) mutiply ,add the corresponding terms you ll get a cubic equation first value of x is found by trial and error method when you get first value of x say x=-4, divide the cubic equation by x+4=0 youll get a quadratic equation solve it and get the values

OpenStudy (anonymous):

shouldnt we do (x-1)(x+2) times 2x-2

OpenStudy (anonymous):

just wait iam typing method

OpenStudy (anonymous):

2x^2+7x-3=-(x+2)/2

OpenStudy (anonymous):

you get equation till here

OpenStudy (anonymous):

4x^2+16x-x-4 (4x-1)(x+4)=0 solve

OpenStudy (anonymous):

get answer

OpenStudy (anonymous):

where did u get 7x-3

OpenStudy (anonymous):

by taking lcm 2x^2+4x+3x-3/(x-1)(x+2)

OpenStudy (anonymous):

the adding get 2x^2+7x-3

OpenStudy (anonymous):

on lhs

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