Question

How do you solve `x/4=2+(x-3)/3`?

Answer 1

`x=-12`

Explanation:

`color(green)(x/4color(red)(xx1)" "=" "[2color(red)(xx1)]+[(x-3)/3color(red)(xx1)])`

I chose the common denominator (bottom number) to be 12

`color(green)(x/4color(red)(xx3/3)" "=" "[2color(red)(xx12/12)]+[(x-3)/3color(red)(xx4/4)])`

`color(green)(" "(3x)/12" "color(white)(.)=" "[24/12]color(white)(.)+" "[(4x-12)/12])`

There are now two ways to think about this.

I love the one that goes: now the denominators are all the same you can forget about them.

The other one is if you are a stickler for mathematical correctness:
Multiply both sides by 12

`color(green)(3x=24+4x-12" "->" "3x=4x+12)`

Subtract `3x` from both sides

`color(green)(0=x+12`

Subtract 12 from both sides

`color(green)(x=-12)`

Answer 2

`"the answer is x=-12"`

Explanation:

`x/4=2+(x-3)/3`

`"let's make equal the denominators at the right side of equation."`

`x/4=color(red)(3/3)*2+(x-3)/3`

`x/4=6/3+(x-3)/3`

`x/4=(6+x-3)/3`

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

`"if "a/b=c/d" then " a*d=b*c`

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

`3x=4*3+4*x`

`3x=12+4x`

`4x-3x=-12`

`x=-12`