Question

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

Answer 1

`x=5`

Explanation:

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

First, we remove the fractions by multiplying the equation by the LCM of `4` and `2`, our two denominators:

LCM of `4` and `2` = `4`

Multiply the equation by `4`.

This means, the expression on the left side of the = sign will be multiplied by `4` and the expression on the right side of the = sign will be multiplied by `4`:

`[(x-3)/4+x/2]color(red)(xx4)=3color(red)(xx4)`

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

`[(x-3)/cancel4xxcancel 4]+[x/cancel2xxcancel4^2]=12`

`x-3+(x xx2)=12`

`x-3+2x=12`

`3x-3=12`

Next, add `3` to both sides of the equation:

`3x-3color(red)(+3)=12color(red)(+3)`

`3x=15`

Finally, divide both sides by `3`:

`(3x)/color(red)(3)=15/color(red)(3)`

`x=5`

You can check your answer by putting back the value `x=5` in the question:

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

Solving the left side:

`= (5-3)/4+5/2`

`= 2/4+5/2`

`=1/2+5/2`

`=(1+5)/2`

`=6/2=color(red)(3)`

Answer 2

x = 5

Explanation:

To eliminate the fractions in this equation `color(blue)"multiply all terms on both sides"` by the L.C.M. (lowest common multiple) of 2 and 4 which is 4.

`rArr[cancel(4)^1 xx(x-3)/cancel(4)^1]+[cancel(4)^2xxx/cancel(2)^1]=4xx3`

The equation now simplifies to

x - 3 + 2x = 12

hence : 3x - 3 =12

and 3x = 15

divide both sides by 3

`(cancel(3)^1 x)/cancel(3)^1 =15/3`

`rArrx=5" is the solution"`