Question

How do you solve `(3w+8)/2=25`?

Answer 1

w = 14

Explanation:

`(3w + 8)/2 = 25`

`3w + 8 = 50`

`3w = 50 - 8 = 42`

Hence `w = 42/3 = 14`

Answer 2

`w=42/3 = 14`

Explanation:

`color(magenta)("With practice you will be able to answer this question type in 1, 2 or 3 lines")`

Gradually you manipulate the equation until you have just a single `w` and it is on 1 side of = and everything else on the other side.

For this equation there are three basic principles that are the `ul("seed principles behind the shortcut methods")`

`color(brown)("Principle 1")`
To maintain the 'truth' of the equation (some people use the word balance): what you do to one side of the equation you do to the other.

`color(brown)("Principle 2")`
To 'get rid' of an add or subtract change it to 0

`color(brown)("Principle 3")`
To get rid of a multiply or divide change it to 1

Note that LHS is left hand side and that RHS is right hand side.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:`" "color(brown)((3w+8)/2=25)`

Using principle 3 to 'get rid' of the denominator of 2 on the LHS
Multiply both sides by `color(blue)(2)`

`color(brown)((color(blue)(2))/2xx(3w+8)=color(blue)(2xx)25)color(green)(" " ->" " 3w+8=50)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using principle 2 to 'get rid of the 8 on the LHS

Subtract `color(blue)(8)` from both sides

`color(brown)(3w+8color(blue)(-8)=50color(blue)(-8)color(green)(" "->" "3w=42)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using principle 3 to 'get rid of the 3 on the LHS

Divide both sides by `color(blue)(3)` to 'get rid' of the 3 on the LHS

`color(brown)(3/(color(blue)(3)) w=42/(color(blue)(3))color(green)(" "->" "w=42/3)`