Question

How do you solve `3x - 2/7 = (3x)/4 + 4`?

Answer 1

`x=40/21`

Explanation:

`3x-2/7=(3x)/4+4`

Make each term have a denominator of `7` on the left side and `4` on the right side.

`(7(3x))/7-2/7=(3x)/4+(4(4))/4`

`(21x)/7-2/7=(3x)/4+16/4`

Subtract the fractions on the left side and add on the right side.

`(21x-2)/7=(3x+16)/4`

Cross multiply.

`7(3x+16)=4(21x-2)`

Multiply.

`21x+112=84x-8`

Isolate for `x` by bringing all terms with `x` to the left and all without to the right.

`21x-84x=-8-112`

Solve.

`-63x=-120`

`x=(-120)/-63`

`x=40/21`

Answer 2

`color(green)(x= 40/21)`
The solution looks a bit long. This is because I have explained the principles behind the shortcuts usually adopted.

Explanation:

`color(blue)("Objective")`
To have a single `x` on the left of the equals sign and everything else on the others side.

`color(blue)("Principles used")`

To end up with something on its own you have to 'remove' from that side the things you do not wish to be there.

For conditions of add or subtract you change what you do not need into the value of 0. This because adding or subtracting 0 has no effect.

For conditions of multiply or divide you change what you do not want into 1. Multiplying or dividing by 1 has now effect.

What you do to one side of an equation you do to the other to maintain the truth od the = sign.

`color(blue)("Application of principle")`
Given: `color(white)("...")color(brown)(3x-2/7=(3x)/4+4..........(1))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Moving `(3x)/4` from the right to the left: subtract `color(blue)((3x)/4)` from both sides of equation (1)

`color(brown)((3x-2/7)color(blue)(-(3x)/4) =((3x)/4+4)color(blue)(-(3x)/4))`
`color(white)(..)`

`color(brown)(3xcolor(blue)(-(3x)/4)-2/7=4+(3x)/4color(blue)(-(3x)/4)` `color(white)(..)`

`color(brown)(3xcolor(blue)(-(3x)/4)-2/7=4+0..................(2)`
`color(white)(..)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But `color(brown)(3x)=(12x)/4color(white)(..)` so `color(white)(..)color(brown)(3x)color(blue)(-(3x)/4)=(12x-3x)/4 =color(purple)((9x)/4)`
So equation (2) becomes:

`color(brown)(color(purple)((9x)/4)-2/7=4.....................(3)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
By applying the same process to equation (3) we have

`(9x)/4= (28+2)/7`

Multiply both sides by `color(blue)(4)` giving:

`color(brown)(color(blue)(4xx)((9x)/4)=color(blue)(4xx)(30/7)`

`color(brown)(color(blue)(4)/4xx9x=(color(blue)(4)xx30)/7`

`9x=120/7`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Divide both sides by `color(blue)(9)` ( same as multiply by `color(blue)(1/9)` )

`color(brown)(color(blue)(1/9xx) 9x=color(blue)(1/9 xx)120/7)`

`color(brown)(9/(color(blue)(9)) xx x=120/(color(blue)(9xx)7)`

`x= 120/63`

But `120/63 = 40/21`

`color(green)(x= 40/21)`