Question

Solve the equation and show the check of the potential answer(s). If any answers are excluded values, state this on your answer sheet.
a2 + 4a - 5/a2 + 2a - a + 3/a + 2 =1/a2 + 2a

Can someone please show me how to do this problem with checks? I have submitted it to my homework three times and cannot get the answer correct.

Answer 1

Can't help, I have no idea what the expression is. Is it this?
(a^2 + 4a - 5)/(a^2 + 2a - a) + 3/(a + 2 )=1/(a^2 + 2a )

Answer 2

Yes that's the one Bob

Answer 3

(a^2 + 4a - 5)/(a^2 + 2a - a) + 3/(a + 2 )=1/(a^2 + 2a )
(a+5)(a-1)/( a^2+2a-a) + 3/(a+2) = 1/a(a+2)

I Suspect at this point, you do not have the problem correct, the (a^2+2a-a) term looks suspiciously wrong

Answer 4

Looking at the distribution of denominators of (a-1), (a+2), a, and another (a+2), is suspect that mysterious denominator is (a+2)(a-1) or (a^2 + a - 2)

so
(a+5)(a-1)/( a^2 + a -2) + 3/(a+2) = 1/a(a+2)
(a+5)(a-1)/((a + 2)(a - 1) + 3/(a+2) = 1/(a(a+2) )
(a+5)/(a + 2) + 3/(a+2) = 1/a(a+2)
multiply each term by a(a+2)
a(a+5) + 3a = 1
a^2 + 5a + 3a - 1 = 0
a^2 + 8a = 1
completing the square:
a^2 + 8a + 16 = 1 + 16
(a+4)^2 = 17
a+4 = ± √17
a = -4 ± √17 <------ based on my assumption