Solve the following equation: \(\frac{2}{(2r - 1)}\) - \(\frac{5}{3}\) = \(\frac{1}{(r + 2)}\)
( -1,\(\frac{5}{2}\) )
( 1, - \(\frac{5}{2}\) )
( \(\frac{5}{2}\), 1 )
(2,1)
Explanation
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Discussions (11)
The answer is correct, but the working is incorrect
2r + 4 - 2r + 15 5
_______________ = ___
2r² +3r - 2 3
Cross multiply
5(2r²+3r-2) = 3(4+1)
10r²+15r-10=15
Factorize
10r²+15r-10-15 = 0
10r²-15r-25=0
(Two numbers that when multiplied give 250, and when added or sub. gives 15 = 10 and 25)
10r²-10r+25r-25 =0
10r(r-1) + 25(r-1) = 0
(r-1) (10r+25) = 0
:- either:
r-1 = 0 or 10r+25 = 0
r =0+1 or 10r = 0-25
r= 1 or r = -25/10
r = 1 or r = -5/2
hit the like button if you agreed with me.
2r + 4 - 2r + 15. 5
_______________ = ___
2r² +3r - 2. 3
cross multiply........ 
Lol
How did they want someone to solve this within a space of 30 sec to minute
Jamb Jamb, una no do well oo
The number '5' which is the numerator in the fraction 5/3 wasn't used in the equation, which i think is wrong. No offense pls
