Given the equation #6=-s+77#, what is the value of #1+5(77-s)#?

2 Answers
Oct 22, 2016

We want to first figure out what the value of #s# is, so let's do that.

If #6=-s+77#, we can rearrange the formula to get #s=71#:

#6=-s+77#
#6+s=77#
#s=71#

Now that we know the value of #s#, we can substitute it into the formula #1+5(77-s)#.

Let's sub in #s=71# and solve it.

#1+5(77-71)#
#1+5(6)#
#1+30 =31#

Oct 22, 2016

#1+5(77-s)=31#

Explanation:

The trick to reducing the solution work for this question is spotting that one is a part repeat of the other.

Let the unknown value be #x# then

#x=1+5(77-s) ............Equation(1)#
#6=-s+77.................Equation(2)#

Change the order of equation(2) giving

#x=1+5(77-s)color(white)(.) ..........Equation(1)#
#6=77-s color(white)(.)......................Equation(2_a)color(red)(" "larr"part repeat"#

Substitute for #77-s# in Equation(1) using #Equation(2_a)#

So Equation(1) becomes:

#x=1+5(77-s)" "->" "x=1+5(6)#

#x=31#