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"#