What exactly did you try?
Given,
y = f(x) = (3x - 5) / (2x - m)
Find the value of m such that f(y) = x
I tried doing it myself, but it just kept getting lengthier & didn't seem to end..
Any idea how to solve that?
Thanks for your time.
This is what I did: -
If y = f(x) = (3x - 5) / (2x - m) ----- (1)
then, f(y) = (3y - 5) / (2y - m) = x [given condition]
or, [3 {(3x - 5) / (2x - m)} - 5] / [2 {(3x - 5) / (2x - m)}] = x [putting the value of y from eqn. (1)]
then....
this is where I came: -
(-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x
This is where I got stuck & couldn't farther anymore steps..
Thanks for the correction.
I did do , and not
Because only after doing , can you arrive at (-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x.
It's just that I forgot to write it here as I'm new to this web.
However, could you just give a simple example to show how to equate the powers of X on both sides of an equation?
I know I learned that at Grade 8 or 9.. But I don't remember the exact procedure.
So if you could just give a simple example of it, that would be very helpful..
Thanks for your time
No problem. In your equation, which looks correct, you can scan the whole equation for the power x^2, and equate all those powers. On the LHS, you don't have any. On the RHS, you have 6-2m. Therefore, 0 = 6-2m.
Next, you can equate the single powers of x^1. On the LHS, you have -1, and on the RHS, you have -10+m^2. Hence, you have the equation -1 = -10 + m^2.
Finally, you can equate the constants (powers of x^0). On the LHS, you have 5m-15, and on the RHS, you have 0. Hence, 5m-15 = 0.
Thankfully, or perhaps the problem was designed with this in view, there is only one value of m that satisfies all three equations, but it does satisfy all three equations. You'd certainly never expect there to be a solution, since over-determined systems of equations often don't have solutions. However, this system does.
Make sense?