Inverse Functions Problem?

Question

y = 1/5 x +b and y = ax+3 are inverse functions.
Solve for a and b...?
Yes I know it's late at night, but I'm stuck on this ONE problem and I can't go to sleep without solving it! Please try your best to give a clear explanation!
And only serious answers please!
Thank you. (Clearest answer gets 10 points! <3)

oh yea, and I need to solve for the *values* of a and b...not just in terms of x or y.

Answer 1

Lets do some relabling:

g = 1/5 y +b and y = ax+3 are inverse functions.

When one function becomes an argument for its inverse, it has to produce the independent argument,x.

x = g(y(x))

x = 1/5 (a x +3) +b <---- property of inverse functions

x - 1/5 a x = 3/5 + b <--- this has to hold for any x, so

the coefficient in front of x and the free term should vanish

a = 5
b = -3/5

if you plot the lines

y = 1/5 x - 3/5
and
y = 5 x + 3

they should be symmetrical with respect to the line y=x,
as it should be for inverse functions(another eg, y=x^2 and y=sqrt(x) )

homework? -- plug g into y (instead of y into g as I did above)
and you should get the same for a and b:

x = a (1/5 x + b) + 3

Answer 2

y = ax + 3

Lets find x: y-3 = ax

So (accepting that a is not 0):

x = (y-3)/a = y/a - 3/a

Lets now change x and y

y = x/a -3/a

Since y = (1/5) x + b = x/5 + b, then I know that a = 5 [x/a is really x/5]

So: y = x/5 + b = x/5 - 3/a

So: b = 3/5 [a=5]

Ana

Answer 3

Since they are inverse functions then you should reverse that so y=ax+3 becomes (x-3)/a. then you set them equal (x-3)/a=1/5x+bthen you solve for each variable or a=(x-3)/(1/5x+b) and B=(x-3)/a-1/5x=b hopefully this will help.

Answer 4

It looks like a = 1/5 and b = 3 makes the two equations identical.

Answer 5

commencing with y =6/(3x-5) or (3x-5)y = 6 (3y-5)x = 6 3y- 5 = 6/x 3y = 6/x +5 y = 2/x+5/3 i'd desire to admit I did it three times earlier I have been given the algebra top. the two graphs are very exciting to verify on the calculator. i've got faith you call those rational purposes. FYI

Answer 6

i am only in middle school, but it doesn't mean i'm not able to solve this one!

-1/5x+y=b (y-3=a)/x
-x+5y=5b
(y-3=a)/x

-x2 + 5xy=5by

y-3=a

a=y-3

(-x2 + 5xy)/5y=b


answers:
a=y-3
b= (5xy - x2)/5y *

*when the step is in parentheses, it means that the whole part in the parentheses is over the other number thingy.

it's really simple.

oh...you want in actual NUMBERS.....

since a=y-3, y=3

a=3

(15x-x2)/15

x(15-x)/15

um....um....
the farthest i went was to Quadratic equations and complex numbers, okay?what do you expect? i'm 12!!!

answer:
a=3
b=x(15-x)/15 because it is impossible to solve for x unless i made a mistake. i'm too lazy to check, okay?