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

3 Answers
Mar 16, 2017

Answer:

#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.

Mar 16, 2017

Answer:

#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"#

Mar 16, 2017

Answer:

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#