Question

How do you solve `-2- \frac { 1} { 2} f = \frac { 1} { 3} f`?

Answer 1

`f=-12/5`

Explanation:

Add `1/2f` to both sides

`" "-2=1/3f+1/2f`

Write as: `" "-2=f/3+f/2`

Multiply by 1 and you do not change a value. However, 1 comes in many forma so you can change the way something looks without changing its actual value.

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

`" "color(green)(-2=[f/3color(red)(xx2/2)]+[f/2color(red)(xx3/3)]`

`" "color(green)(-2=" "[(2f)/6]color(white)(.)+color(white)(.)[(3f)/6]`

`" "color(green)(-2=(2f+3f)/6)`

`" "color(green)(-2=5/6f`

Multiply both sides by `6/5`

`" "color(green)(-2xx6/5=5/6xx6/5xxf)`

`" "color(green)(-2xx6/5=5/5xx6/6xxf)`

but `5/5=1" and "6/6=1`

`" "color(green)(-12/5=f)`

Answer 2

`f=-12/5`

Explanation:

To eliminate the fractions in the equation, multiply ALL terms on both sides by the `color(blue)"lowest common multiple"` (LCM) of 2 and 3, the denominators.

The LCM of 2 and 3 is 6

`rArr(6xx-2)-(cancel(6)^3xxf/cancel(2)^1)=(cancel(6)^2xxf/cancel(3)^1)`

`rArr-12-3f=2flarrcolor(red)" no fractions"`

add 3f to both sides.

`-12cancel(-3f)cancel(+3f)=2f+3f`

`rArr5f=-12`

To solve for f, divide both sides by 5

`(cancel(5) f)/cancel(5)=(-12)/5`

`rArrf=-12/5`

`color(blue)"As a check"`

Substitute this value into the equation and if the left side equals the right side then it is the solution.

`"left side "=-2-(1/2xx-12/5)=-2+6/5`

`color(white)(00000000)=-10/5+6/5=-4/5`

`"right side "=1/3xx-12/5=-4/5`

`rArrf=-12/5" is the solution"`