SOLUTION: (5a-4)= a+41/a-1 Please help if you can.

Question 502181: (5a-4)= a+41/a-1
Please help if you can.
Answer by persian52(161) (Show Source): You can put this solution on YOUR website!
(5a-4)=a+(41)/(a)-1
Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
a+(41)/(a)-1=(5a-4)
Find the LCD (least common denominator) of a+(41)/(a)-1+(5a-4).
Least common denominator: a
Multiply each term in the equation by a in order to remove all the denominators from the equation.
a*a+(41)/(a)*a-1*a=(5a-4)*a
Simplify the left-hand side of the equation by canceling the common terms.
a^(2)-a+41=(5a-4)*a
Simplify the right-hand side of the equation by multiplying out all the terms.
a^(2)-a+41=5a^(2)-4a
Move all terms not containing a to the right-hand side of the equation.
a^(2)-a+41-5a^(2)+4a=0
Since a^(2) and -5a^(2) are like terms, add -5a^(2) to a^(2) to get -4a^(2).
-4a^(2)-a+41+4a=0
Since -a and 4a are like terms, subtract 4a from -a to get 3a.
-4a^(2)+3a+41=0
Multiply each term in the equation by -1.
-4a^(2)*-1+3a*-1+41*-1=0*-1
Simplify the left-hand side of the equation by multiplying out all the terms.
4a^(2)-3a-41=0*-1
Multiply 0 by -1 to get 0.
4a^(2)-3a-41=0
Use the quadratic formula to find the solutions. In this case, the values are a=4, b=-3, and c=-41.
a=(-b\~(b^(2)-4ac))/(2a) where aa^(2)+ba+c=0
Substitute in the values of a=4, b=-3, and c=-41.
a=(-(-3)\~((-3)^(2)-4(4)(-41)))/(2(4))
Multiply -1 by the -3 inside the parentheses.
a=(3\~((-3)^(2)-4(4)(-41)))/(2(4))
Simplify the section inside the radical.
a=(3\~(665))/(2(4))
Simplify the denominator of the quadratic formula.
a=(3\~(665))/(8)
First, solve the + portion of \.
a=(3+~(665))/(8)
Next, solve the - portion of \.
a=(3-~(665))/(8)
The final answer is the combination of both solutions.
a=(3+~(665))/(8),(3-~(665))/(8)_aAPPR3.6,-2.85
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