Question

How do you solve `7( 3s + 4) = 301`?

Answer 1

`s=13`

Explanation:

`7(3s+4)=301`

First multiply out the brackets:

`7xx3s+7xx4=301`

`21s+28=301`

Next subtract `28` from both sides of the `=` sign:

`21s+28-28=301-28`

`21s=273`

Next divide both sides by 21 and reduce:

`(cancel21^color(red)(1)s)/(cancel21^color(red)1)=(cancel273^color(red)13)/(cancel21^color(red)1)`

`s=13`

It is always a good idea (and very simple) to check your answer when you've finished.

Replace the constant with the value you found in the original formula. So in this case:

`7(3xx13+4)=301`

`7(39+4)=301`

`7xx43=301`

`301=301`

The left hand side (LHS) and right hand side (RHS) of the equal's match, proving the answer correct.

Answer 2

`s=13`

Explanation:

The first step is to distribute the bracket on the left side.

`rArr21s+28=301`

subtract 28 from both sides.

`21scancel(+28)cancel(-28)=301-28`

`rArr21s=273`

divide both sides by 21

`(cancel(21) s)/cancel(21)=273/21`

`rArrs=13" is the solution"`

Answer 3

Just another approach!

`s=13`

Explanation:

The objective is to work you way to getting just 1 of s and for it to be on its own on one side of the = and everything else on the other side.

Isolating the `ul("'group'")` of values that has s in it- divide both sides by 7

`3s+4=301/7`

Isolating 3s: subtract 4 from both sides

`3s=301/7-4`

Final isolation step: divide both sides by 3

`s=301/(3xx7)-4/3`

`s=301/21-4/3`

But `3xx7=21`

`color(green)(s=301/21-[4/3color(red)(xx1) ])`

`color(green)(s=301/21-[4/3color(red)(xx7/7) ])`

`s=301/21-28/21`

`s=273/21 = 13`