DOES ANYONE KNOW HOW TO SOLVE INVERSES ON myalgebra.com?

I'm not sure how to solve inverses on the website, but it's really easy if you look closely.

g(x)=(insert x value here) is the same as saying y=(insert x value here.

So, saying g(x)=5x+5 is the same as saying y=5x=5.

To find the inverse for this equation, you just switch x and y, and then solve for y.

For example:

g(x)=5x+5

y=5x+5

x=5y+5

x-5=5y

((x-5)/5)=y

y=(x-5)/5 is the inverse (or g(x)=(x-5)/5 is the inverse, if you prefer it that way).

Another example (harder this time):

z(x)=17.5x+(-|5.5x|)-14

y=17.5x+(-|5.5x|)-14

y=17.5x-5.5x-14

y=12.5x-14 (Now the equation is in simplest form)

x=12.5y-14

y=(x+14)/12.5

z(x)=(x+14)/12.5 is the inverse.

If you need more help, feel free to shoot me a message. :)

g(x) = 5x + 5

y = 5x + 5

Switching y and x

x = 5y + 5

5y = x - 5

y = (x - 5)/5

g^-1(x) = (x - 5)/5

The example

g(x) = 17.5x + (-|5.5x|) - 14

y = 17.5x + (-|5.5x|) - 14

= 17.5x - √ ̅(5.5x)^2 - 14

√ ̅(5.5x)^2 = 17.5x - 14 - y

(5.5x)^2 = (17.5x - 14 - y)^2

Switching x and y

(5.5y)^2 = (17.5y - 14 - x)^2

5.5y = ±√ ̅(17.5y - 14 - x)^2

For +

5.5y = 17.5y - 14 - x

y = (x + 14)/12

For -

5.5y = -(17.5y - 14 - x)

y = -(x + 14)/23

So

g^-1(x) = (x + 14)/12 or

g^-1(x) = -(x + 14)/23