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

##### 3 Answers

#### Answer:

#### Explanation:

First multiply out the brackets:

#7xx3s+7xx4=301#

#21s+28=301#

Next subtract

#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:

#### 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:

Just another approach!

#### 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

Isolating 3s: subtract 4 from both sides

Final isolation step: divide both sides by 3

But