Question #97f62

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Answer 1 · 2017-05-04T15:56:27

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

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