distribute 4

4f-12=4f

subtract 4f from both sides.

you get -12=0

this means, there is no solution to f.

The equation breaks down to : f – 3 = f by dividing both sides by four

then

– 3 = f – f which doesn’t make sense because – 3 obviously does not equal 0, so faced with an impossibility we then think well instead of looking at absolute sums what would happen if we put the equation on a number line instead? A number line place the equation back at -3 = +f-f. Since negative numbers are a pain I’m going to multipy both sides by -1

so :: +3 = -f +f

Then you can picture in your mind an answer that is not impossibility. And still keep the values of 3=0 (originally it was -3=0 but I multiplied by -1).

on a number line the ” 0 ” would be in the middle which fits the criteria we need above a number that is 0 but yet at the same time is not 0

When we do that then -f and +f does have a value on the number line

– 5 -4 -3 – 2 -1 0 +1 +2 +3 +4 +5

We already accounted for the 0 but what about the negative 3 ? (again temporarily changed to positive 3 because it makes it easier to see the graph in your head).

we need a number which is equal to 3 whole places on the number line and 0 is in the middle.

so f = negative 1 and 1/2 or positive 1 and 1/2.

So when placed on a number line -3 does equal 0 in the confines of the original equation.

ps if I remember my math correctly the answer is written like this : f = (-1 1/2, +1 1/2) or f = ( -1.5,+1.5) where the solution is said to be a “set” of numbers instead of having just one absolute answer.

4f – 12 = 4f

-12 = 0

either you have the equation wrong …. or you were given an equation with NO SOLUTION