Getting Rid of 1 over a Variable

[Jason76]

I know you can cross-multiply on this one. But let's say we're going a different route (which this is good to know for problems for "1 over a variable" where you can't cross multiply):

\(\displaystyle \dfrac{50}{5} = \dfrac{40}{y}\)

\(\displaystyle \dfrac{50}{40(5)} = \dfrac{1}{y}\) - Take reciprocal of both sides to get rid of "1 over y"

\(\displaystyle \dfrac{40(5)}{50} = y\)

\(\displaystyle \dfrac{40(5)}{50} = y\)

\(\displaystyle 4 = y\)

But what if

\(\displaystyle \dfrac{-50}{5} = \dfrac{40}{y}\)

\(\displaystyle \dfrac{-50}{40(5)} = \dfrac{1}{y}\) - If you take the reciprocal of both sides, then does the negative sign on the 50 change?

\(\displaystyle \dfrac{40(5)}{-50} = y\)

\(\displaystyle \dfrac{40(5)}{-50} = y\)

\(\displaystyle -4 = y\)

 

[Jason76]

Sure you can : 1/a = 2/b ; 1*b = 2*a

You can criss cross multiply also for "a variable over 1":
a/1 = 2/b ; a*b = 1*2

Sure, but in some cases we want an equation in the form: "\(\displaystyle y =\) other part of function". Not "\(\displaystyle \dfrac{1}{y} =\) other part of function". Some situations it's impossible to get it in the right form using crisscross multiplying. Am I wrong?

 

[HallsofIvy]

I can't imagine a situation where you can "take the reciprocal of both sides" but can't "cross multiply". Can you give an example?

 

[JeffM]

Sure, but in some cases we want an equation in the form: "\(\displaystyle y =\) other part of function". Not "\(\displaystyle \dfrac{1}{y} =\) other part of function". Some situations it's impossible to get it in the right form using crisscross multiplying. Am I wrong? YES, you are wrong.

\(\displaystyle \dfrac{a}{b} = \dfrac{c}{d} \implies\)

\(\displaystyle b * \dfrac{a}{b} = b * \dfrac{c}{d} \implies\)

\(\displaystyle a = \dfrac{bc}{d} \implies\)

\(\displaystyle a * d = \dfrac{bc}{d} * d \implies\)

\(\displaystyle ad = bc.\)

In short, \(\displaystyle \dfrac{a}{b} = \dfrac{c}{d} \implies ad = bc.\)

\(\displaystyle However,\ it\ is\ false\ that\ ad = bc \implies \dfrac{a}{b} = \dfrac{c}{d}.\)

The latter proposition may be what you are thinking about.

 

[mmm4444bot]

Some situations it's impossible to get it in the right form using crisscross multiplying. Am I wrong?

I'm not sure what you may be thinking, here. For example, are you thinking that one ought to be able to get the desired form in a single step, by using cross-multiplication? That's not always the case.

I'd like to see the situation(s) which you have in mind. :cool: