How do you use cross products to solve `3/4=x/(x+3)`?

`3/4=x/(x+3)`
or
`4x=3x+9`
or
`4x-3x=9`
or
`x=9`

To use cross products or `color(blue)"cross multiplication"` as it is also named.

`color(red)(3)/color(blue)(4)=color(blue)(x)/color(red)(x+3)`

now multiply the terms in `color(blue)("blue")" and "color(red)("red")` (X) and equate them.

`rArrcolor(blue)(4x)=color(red)(3(x+3))`

distribute the bracket : 4x = 3x + 9

subtract 3x from both sides to solve for x

`4x-3x=cancel(3x)+9cancel(-3x)rArrx=9`

The cross product is a shortcut that bypasses some stages in solving by first principles. I will use first principles so you can see where the shortcut takes over.

A fraction is split up into two parts. Using descriptive but `ul("unconventional names")` we have `("count")/("size indicator")`

When you wish to `ul(directly")` compare quantities the "size indicators" have to be the same. This is also true for fractional addition and subtraction. You can not `ul("directly")` apply addition or subtraction unless the "size indicators" are the same.

`" "("count")/("size indicator") ->("numerator")/("denominator")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving your question")`

Using a 'common denominator' of `4(x+3)`

`color(brown)([3/4xx1]=[x/(x+3)xx1]color(blue)(""->""[3/4xx(x+3)/(x+3) ]=[x/(x+3)xx4/4]`

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

If you look at the numerators you will see the result you get by the short cut

Multiply both sides by `4(x+3)` and you end up with

`3(x+3)=4x" "larr" the consequences of the shortcut"`