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June 27, 2016
Posted by **Shreya** on Tuesday, January 22, 2013 at 9:18pm.

I got to this step

x+4 = 2/3y

I need to divide by 2/3 to get rid of 2/3 beside the y, but i dont know how to divide x+4/2/3 that looks complicated :\

- Pre Calculus 12 -
**Reiny**, Tuesday, January 22, 2013 at 9:38pmHere is the way I do these:

y = (2/3)x - 4

to find the inverse, I follow these steps

1. to find the inverse equation, simply interchange the x and y variables

So the inverse of y = (2/3)x - 4 is

x = (2/3)y - 4

2. Now you want to probably solve this for y

so multiply each term by 3

3x = 2y - 12

3x + 12 = 2y**y = (3/2)x + 6**

to test it, take any value of x in the original

say x = 9,

y = (2/3)(9) - 4 = 2

now put that in as your x value, x = 2 in the inverse

y = (3/2)(2) + 6 = 9 , as expected

some folks clear the fraction first, then interchange the x and y variables, that is your choice

Try it , you must get the same answer - Pre Calculus 12 -
**Shreya**, Tuesday, January 22, 2013 at 10:03pmthanks Reiny but the answer says y=3/2(x+4) is it supposed to be 6?

- Pre Calculus 12 -
**Reiny**, Tuesday, January 22, 2013 at 10:09pmmmmh, why don't we expand their answer

y = (3/2)(x+4)

= .....

well, what do you know ??

Either answer is correct. - Pre Calculus 12 -
**Shreya**, Tuesday, January 22, 2013 at 10:22pmFor the second step, to get rid of a fraction on the bottom, we always multiply it with every term? And if we are dividing something, then we divide it by every term? K i get how you got 6, but I still dont get how to get the x+4 ?

- Pre Calculus 12 -
**Reiny**, Tuesday, January 22, 2013 at 10:35pmy = (3/2)(x+4)

= (3/2)x + (3/2)(4)

= (3/2)x + 6

showing that my answer is the same as theirs

they simply factored it, there was no need to do that.

If you ever get an answer which looks different from the text book answer, try using proper algebraic manipulation to see if they are the same

Another way is to sub in some value of x, (like I did above) into both equations. If you get the same value, the two equations are equivalent.

And yes, if you multiply or divide one term by some number , you must multiply or divide each term by that same number

Remember, "Whatever you do to one side of an equation, you must do to the other side" - Pre Calculus 12 -
**Shreya**, Tuesday, January 22, 2013 at 10:48pmThanks Reiny that makes soo much more sense :)