
Simple algebra
Hi this is a very easy question but for some reason, I cannot identify my error..
Xin and Yu opened a business in 2001. Their annual profit was 160000 pounds in 2004, 198 000 pounds in 2005 and 240 000 pounds in 2006. Based on the info above, they believe that their annual profit can be predicted by the model,
P(t) = at +b + $\displaystyle \frac{c}{t+4}$ pounds
where t is the number of years after 2004, ie. t=0 gives the profit for 2004.
Determine the values for a,b,c which fit the profits for 2004, 2005 and 2006.
For 2004, I got b + $\displaystyle \frac{c}{4} = 160 000$, 4b+c = 640 000 and therefore c = 640 0004b
For 2005, I got a+b+$\displaystyle \frac{c}{5} = 198 000$, 5a+5b+c = 990 000, 5a+5b+640 0004b = 990 000, 5a+b = 350 000,
therefore b = 350 0005a
For 2006, 2a+b+$\displaystyle \frac{c}{6} = 240 000$, 12a+6b+c = 1440 000, 12a+6(350 000  5a )+ 640 000  4b = 1440 000, 12a + 2100 000  30a + 60 0000 4(350 0005a) = 1440 000,
2a = 100 000,
a = 50 000
b = 150 000
c = 1240 000
Unfortunately, only my answer for a was right. I can't seem to spot my mistake..
Thank you very much for your time!

Re: Simple algebra
Looks like just a computational mistake.
You got a=50000
b=3500005a= 3500005(50000)= 100000
c = 640 0004b= 6400004(100000)= 240000