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Can someone help me solve that?
I've been racking my brain for hours. D: It's so hard.
I've tried factoring and cancelling out the denominators, but then I get seriously stuck with all the factors.
2007-04-09 20:10:16 · 6 answers · asked by eqlar 3 in Science & Mathematics ➔ Mathematics
Factor denominators into:
(4a+1) (2a-1) and (2a-1) (a+3) and (4a+1) (a+3)
Now,
Multiply both sides by (2a-1)
Then multiply both sides by (a+3)
Then multiply both sides by (4a+1)
Now all that are left are numerators
2a^2 +13a +21 + 4a^2-15a -4 = 8a^2 - 6a+1
simplify
6a^2-2a+17 = 8a^2-6a +1
2a^2-4a -16 = 0
a^2-2a-8=0
(a-4)(a+2) =0
a = 4 and -2
2007-04-09 20:29:57 · answer #1 · answered by ignoramus 7 · 1⤊ 0⤋
NOTE
2 after 'a' denotes the power
1] 2a+7 / 8a2-2a-1 can be factorized as
2a+7 / (4a+1)(2a-1)
2] a-4 / 2a2+5a-3 = a-4/(2a-1)(a+3)
3] 4a-1 / 4a2+13a+3 = 4a-1 / (4a+1)(a+3)
adding 1 and 2 by taking LCM
(2a+7)(a+3)+(a-4)(4a+1) / (4a+1)( a+3 )(2a-1)
=4a-1 /(4a+1)(a+3)
cancel ling the denominators and cross multiplying
u get
2a2-4a-16= a2-2a-8=(a-4)(a+2)
2007-04-09 20:52:37 · answer #2 · answered by bullet 2 · 1⤊ 0⤋
answer:
a= -4/3 or a=2
2007-04-09 20:53:53 · answer #3 · answered by eclipse 2 · 0⤊ 0⤋
theres no 9's in the problem it's all a's
2007-04-09 20:28:29 · answer #4 · answered by cocomademoiselle 5 · 0⤊ 0⤋
im not able to make out if its 9 or "a"
2007-04-09 20:21:25 · answer #5 · answered by naughtyme12345 1 · 0⤊ 0⤋
((2a+7)/(8a^2-2a-1))+((a-4)/(2a^2+5a-3))=(4a-1)/(4a^2+13a+3))
=>((2a+7)/((2a-1)(4a+1)))+((a-4)/((2a-1)(a+3)))=((4a-1)/((4a+1)(a+3)))
=>((2a+7)(a+3)+(a-4)(4a+1))/(2a-1)=4a-1
=>2a^2+6a+7a+21+4a^2+a-16a-4=(2a-1)(4a-1)
=>6a^2-2a+17=8a^2-6a+1
=>2a^2-4a-16=0
=>a^2-2a-8=0
=>a^2-4a+2a-8=0
=>a(a-4)+2(a-4)=0
=>(a+2)(a-4)=0
=>a=4,-2
2007-04-09 20:36:32 · answer #6 · answered by Molly 1 · 1⤊ 0⤋