Question

How do you subtract two rational numbers with different denominators?

Answer 1

Generic explanation given using algebra. Followed by a numeric example.

Explanation:

`color(blue)("Initial concept")`
A fractions construct is such that we have:

`("numerator")/("denominator") -> ("count")/("size indicator of what is being counted")`

This may be further considered as:

`" count" color(white)("d")ubrace(" of ")color(white)("ddd")1/("size indicator of what is being counted")`
`color(white)("ddddddd.")darr`

`"count "color(white)("d.")xxcolor(white)("ddd")ubrace(1/("size indicator of what is being counted"))`
`color(white)("dddddddddddddddddddddddddd.")darr`

`"count "color(white)("d.")xxcolor(white)("dddddddddd")" unit of measurment"`

`color(red)("To directly add counts the units of measurement must be the same")`
`color(green)("To directly add numerators the denominators must be the same")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

Let the first rational number be `a/b`
Let the second rational number be `c/g`

Consider the context:
`a/b-c/g larr" denominators are not the same"`

I opted for `g` instead of `d` as its difference in look to `b` if very obvious. `b and d` are easily confused with each other.

Multiply by 1 and you do not change the actual value. However, 1 comes in many forms.

`color(green)(a/b-c/g color(white)("dddd")->color(white)("dddd")[a/bcolor(red)(xx1)]-[c/gcolor(red)(xx1)] )`

`color(green)(color(white)("dddddddddd")->color(white)("dddd")[a/b color(red)(xxg/g)]-[c/g color(red)(xxb/b)] )`

`color(green)(color(white)("dddddddddd")->color(white)("dddddd")[(ag)/(bg)]color(white)("d")-color(white)("d")[(bc)/(bg)] )`

`color(green)(color(white)("dddddddddd")->color(white)("ddddddddd")(ag-bc)/(bg))`

So basically you change the denominators to the same value and then directly subtract the suitably adjusted numerators.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Example")`

`3/5-7/10`

`[3/5xx2/2]-7/10`

`[(3xx2)/(5xx2)]-7/10`

`color(green)(6/10color(red)(-7/10))`

But `color(red)(-7/10)` may be split into `color(red)(-6/10-1/10)` giving:

`color(green)(ubrace(6/10color(red)(-6/10))color(red)(-1/10))`
`color(white)("d.d")darr`
`color(white)("ddd")0color(white)("dd")-1/10`

So `3/5-7/10= -1/10`