inverse equations

Question

can anyone help me find the inverse of these 2 equations?

1. Y = 5/x+2
2. Y= x+1/2+x

Answer 1

Inverse functions are much akin to replacing x->y and y->x and resolving...

1) y = 5/x + 2 --> x = 5/y + 2

x - 2 = 5/y

(x-2)/5 = 1/y

5/(x-2) = y

2) y = x + 1/(2+x) --> x = y + 1/(2+y)

x = y*(2+y)/(2+y) + 1/(2+y) = (2y + y² + 1)/(2+y) = (y+1)²/(y+2)

x = (y+1)²/(y+2)

x*(y+2) = (y+1)²

xy + 2x = (y+1)²

xy + 2x = y² + 2y + 1

-1 + 2x = y² + 2y - xy = y² + (2-x)*y

Now complete the square on the right side of equal sign...

a = 1

b = 2-x

b/2a = (2-x)/2

y² + (2-x)*y = y² + (2-x)*y + [ (2-x)/2 ]² - [ (2-x)/2 ]²

= [ y + (2-x)/2 ]² - [ (2-x)/2 ]²

-1 + 2x = [ y + (2-x)/2 ]² - [ (2-x)/2 ]²

-1 + 2x + [ (2-x)/2 ]² = [ y + (2-x)/2 ]²

Now expand and collect terms on the right hand side...

x + x²/4 = [ y + (2-x)/2 ]²

sqrt(x + x²/4) = [ y + (2-x)/2 ]

y_new = sqrt(x + x²/4) - (2-x)/2

Answer 2

Taylor, please submit this again with parentheses where they need to be.

So, for the first one, is it

a). Y = 5/x +2, which is equivalent to

y = (5/x)+2

or is it

b). 5

y = ———

X + 2

if you want this, you need to write

y = 5/(x+2)

The second problem is similar, as there are a number of ways to parenthesize it, all different.

Please parenthesize accurately, and we can help.