hello math people of basil
Find the values of __a__ and __b__ for which: __a__(x+2) + 2__b__ = (1 + 2x)__b__ + 6
GO!
also, the = was the identical to sign (the equal sign but with 3 bars) if that matters
TYY
April 18, 2015
3 Comments • Newest first
HolyDragon is correct; here's how I do it.
a(x+2) + 2b = (1 + 2x)b + 6
ax + 2a + 2b = b + 2bx + 6
Looking at coefficients of x:
a = 2b (1)
Non coefficients of x:
2a + 2b = b + 6
2a + b - 6 = 0 (2)
Solve (1) and (2) simultaneously by subbing (1) into (2):
2(2b) + b - 6 = 0
4b + b - 6 = 0
5b - 6 = 0
b = 6/5
Sub b=6/5 back into (1)
a = 2*6/5
a = 12/5
https://www.wolframalpha.com/ is pretty useful.
a(x+2) + 2b = (1 + 2x)b + 6
a(x+2) +2b = b +2bx +6
a(x+2) = 2bx -b +6
a(x+2) = b(2x -1) +6
a(x+2) - b(2x-1) -6 = 0
x(a-2b) +(2a+b)-6 =0
Since the variables of x has to be zero
a-2b=0
Since the other numbers also has to be zero
2a+b-6=0
I think I might've done something wrong somewhere. Did not expect to get something definite.