Thread: Can anybody please show me how to solve this problem step by step?

[1/(c-2)] - [3c/(c^2 + 2c - 8)] = -4/ (c+4)
Thanks

Originally Posted by gnarlycarly227

[1/(c-2)] - [3c/(c^2 + 2c - 8)] = -4/ (c+4)
Thanks

You basically want to factorise c^2 + 2c - 8 to (c+4)(c-2) then cancel out.

1/(c-2) = (c+4)/((c-2)(c+4))

-4/(c+4) = -4(c-2)/((c-2)(c+4))

3c/(c^2 + 2c - 8) = 3c/((c-2)(c+4))

then you have....

(c+4)-3c+4(c-2)/((c-2)(c+4)) = 0

2c-4/((c-2)(c+4)) = 0

2(c-2)/((c-2)(c+4)) = 0

2/(c+4) = 0

Wait, did you want to solve or to simplify... oops.

OK, take the line

(c+4)-3c+4(c-2)/((c-2)(c+4)) = 0

(c+4)-3c+4(c-2) = 0

2c-4 = 0

2c = 4

c = 2