Question

How do you solve `\frac { m - 3} { 4} = \frac { m + 1} { 3}`?

Answer 1

`m=-13`

Explanation:

`"multiply both sides by the "color(blue)"lowest common multiple of 4 and 3"" that is 12"`

`cancel(12)^3xx(m-3)/cancel(4)^1=cancel(12)^4xx(m+1)/cancel(3)^1`

`rArr3(m-3)=4(m+1)larrcolor(blue)"distribute"`

`rArr3m-9=4m+4`

`"subtract 4m from both sides"`

`3m-4m-9=cancel(4m)cancel(-4m)+4`

`rArr-m-9=4`

`"add 9 to both sides"`

`-mcancel(-9)cancel(+9)=4+9`

`rArr-m=13`

`"multiply both sides by "-1`

`rArrm=-13`

`color(blue)"As a check"`

Substitute this value into the equation and if both sides are equal then it is the solution.

`"left "=(-13-3)/4=(-16)/4=-4`

`"right "=(-13+1)/3=(-12)/3=-4`

`rArrm=-13" is the solution"`

Answer 2

`m=-13`

Explanation:

In a case where you have:

`"fraction " = " fraction"`,

a quick way of getting rid of the fractions is to cross-multiply.

(This is a short cut based on the result of multiplying both sides by the LCM of the denominators.)

`color(red)((m-3))/color(blue)(4) = color(blue)((m+1))/color(red)(3)`

It does does not matter which pair you multiply first, but notice that in this case, the blue combination will give `4m` while the red will give `3m`. I would choose to multiply the blues first:

`color(blue)(4 xx (m+1) = color(red)(3xx(m-3)`

`4m+4 = 3m-9" "larr`re-arrange and solve

`4m-3m = -9-4`

`m= -13`