Help me letsrun! my brain hurts and this should be easy
b = 1.05(a + 0.925a)
solve for a
Help me letsrun! my brain hurts and this should be easy
b = 1.05(a + 0.925a)
solve for a
1a+0.925a = ?
asdgfwagds wrote:
1a+0.925a = ?
No... that's not what I'm looking for...
I'm hoping to find a = ______b
what's that called? an inverse function?
b = 1.05(a + 0.925a)
b = 1.05a + 0.97125a (I expanded the brackets)
b = a(1.05 + 0.97125) (Factorised for 'a')
b = 2.02125a
a = b/1.02125
P.S: You're shit at maths
newrunner888 wrote:
b = 1.05(a + 0.925a)
b = 1.05a + 0.97125a (I expanded the brackets)
b = a(1.05 + 0.97125) (Factorised for 'a')
b = 2.02125a
a = b/1.02125
P.S: You're shit at maths
Awesome, thanks
you don't even need to do all that :
b = 1.05(a + 0.925a)
-> b = 1.05 ( 1.925a )
-> a = b/2.02125
reminds me of a clever sleight-of-hand problem which impressed me as a kid :
if :
(a + b)^2 = 100
&
a^2 + b^2 = 60
what is the value of :
ab ?
( this problem only for high-schoolers )
any takers ?
I think I got it.
(a+b)^2 = 100
a^2 + 2ab + b^2 = 100 (multiplied out)
-
a^2 + b^2 = 60 ( subtracting the two equations
-2ab = 40
ab = -20
Should be
2ab=40
ab=20
very good
most kids ( i hope you are a kid ) woud try to solve for b or a & then get the product - a method which will get you nowhere fast
subtracting the 2 equations separates men from the boys...
ventolin^3 wrote:
very good
most kids ( i hope you are a kid ) woud try to solve for b or a & then get the product - a method which will get you nowhere fast
subtracting the 2 equations separates men from the boys...
That's how I always solved equations. My classmates thought I was weird. I even got marked wrong in HS once for this since we "weren't taught it yet". I still solved it mathematically! Just because I learned a method in junior high that we hadn't learned yet didn't make it any less correct.
Oh? wrote:
asdgfwagds wrote:1a+0.925a = ?
No... that's not what I'm looking for...
I'm hoping to find a = ______b
what's that called? an inverse function?
No, it IS what you're looking for, I just made the mistake of assuming you wanted help and not the answer.
The first step should be to try to simplify the problem by grouping similar terms.