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Question 27279: (5/n)+(4)/(6-3n) = (2n)/(6-3n)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! (5/n)+(4)/(6-3n) = (2n)/(6-3n)--------(1)
The LCM(Least Common Multiple) of n and (6-3n) is n(6-3n)
Multiplying (1) through out by n(6-3n) we get
5(6-3n) + 4n = (2n)X n
30 - 15n + 4n = 2n^2
30 - 11n = 2n^2
2n^2 = 30 - 11n
(Changing LHS to RHS and RHS to LHS does not call for changing sign and only if one or more terms from a side transferred to another then should change sign)
2n^2 + 11n - 30 = 0 -----(I)(bringing all the terms from the present RHS to LHS)
Factorising, since 2 X (-30) = -60 and the factors of 60 are 2,2,3 and 5,
grouping the factors into 15 and 4 whose difference is 11 the middle term
we write (I) as
2n^2 + (15n - 4n) - 30 = 0
(2n^2 + 15n) - 4n - 30 = 0
n(2n + 15)- 2(2n + 15) = 0
(2n+15)(n-2) = 0
2n + 15 = 0 gives 2n = -15 which means n = -15/2
and (n-2) = 0 gives n = 2
But then since (6 - 3n) = 6 - 6 = 0 for n = 2 and as division by 0 is not allowed, the value 2 for n is not valid
Why did we get it then?
It is because when we processed (1) by multiplying by n(6-3n)
[which means n cannot be zero and n cannot be 2]
we got a quadratic equation in n and any equation of degree k should give k and k alone values our quadratic gave 2 values.
Answer: n = -15/2
Verification:
Putting n = -15/2 in (1)
LHS = (5/n)+(4)/(6-3n) = 5 divided by (-15/2 ) + 4 divided by [6 - 3X(-15/2)]
= 5 X (-2/15) + 4/(6 + 45/2)
= -2/3 + 4 X (2/57)=
= -2/3 + 8/57
(-2/3)X(57/57) + (8/57)
(multiplying numerator and denominator by the lcm 57 of 3 and 57)
= -38/57 + 8/57 = (-38+8)/57 = -30/57 = -10/19 ----*
RHS = 2n)/(6-3n)
= [2 multiplied by (-15/2 )] and divided by [6 - 3X(-15/2)]
= -15/(6 + 45/2)= -15 divided by (57/2)
= -15 X (2/57) = -5 X (2/19) = -10/19 ----**
From * and ** LHS = RHS and hence n= -15/2 holds
A word about factorization.
After multiplying the coefficient of the square term and the constant term,we must observe the sign.
If the sign of the product is postive, and the sign of the middle term is also positive, we must try to group the factors of the product into two sets so as to form their sum equal to the middle term (see to it that all the factors of the product are included without exception)and give them positive sign.( so that positive multiplied by positive is positive and positive added to positive is positive)
If the sign of the product is postive, and the sign of the middle term is negative, we must try to group the factors of the product into two sets so as to form their sum equal to the middle term (see to it that all the factors of the product are included without exception)and give them negative sign.( so that negative multiplied by negative is positive and negative added to negative is negative)
If the sign of the product is negative, and the sign of the middle term is postitive, we must try to group the factors of the product into two sets so as to form their difference equal to the middle term (see to it that all the factors of the product are included without exception)and give the bigger numerical one the sign of the middle term that is positive and the other the negative sign.( so that postitive multiplied by negative is negative and positive added to negative is postive since the larger numerical number carries positive sign)
If the sign of the product is negative, and the sign of the middle term is negative, we must try to group the factors of the product into two sets so as to form their difference equal to the middle term (see to it that all the factors of the product are included without exception)and give them give the bigger numerical one the sign of the middle term that is negative and the other the positive sign.(so that negative multiplied by positive is negative and negative added to positive is negative since the larger numerical number carries negative sign)
Note: When two numbers of opposite signs are added, we give the sign of the bigger number and then take away the smaller from the bigger.
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