Question

Question #97f62

Answer 1

For 1 gallon you have `1/2` of the something else

Explanation:

Assumption: the given ratio is of 2 gallons : 1 of something else
and that you require to determine how much there is 'something else for 1 gallon.

Let 'something else be represent by `s`

Express initial condition ratio in fractional form `(2g)/(1s)`

Let the unknown count or volume of `s` be `x`

`(2g)/(1s)-=(1g)/(xs)`

Turn the whole thing upside down

`(1s)/(2g)-=(xs)/(1g)`

Multiply both sides by `1g larr` gets the `xs` on its own

`color(brown)((1color(green)(s))/(2color(green)(g))xx1color(green)(g)-=xcolor(green)(s))`

Separating numbers from units of count/measurement:

`color(brown)(1/2 xx 1" ")color(green)(s/(cancel(g)) xx cancel(g))`

Numbers left `1/2" units left"-> s`

So for 1 gallon you have `1/2` of the something else