Question #a8045

2 Answers
May 24, 2017

#-40°#

Explanation:

We know #F=9/5C+32#
We need to find a point in which #F=C#
So, we let #F=C# and rewrite the equation to look like this:
#C=9/5C+32#

From here, we just solve for C. Now, in a many cases, when we substitute #F# with #C#, there is no solution and we get something that looks like #1=2#. This means that there is no point in which #F=C#

Let's solve this.
#C=9/5C+32#
#C-9/5C=32#

Factorize #C# out.

#C(1-9/5)=32#
#-4/5C=32#
#C=32*-5/4=-40#

#:.# When #C=-40# #F# is also #-40#

We can check:
#F=9/5C+32#
#-40=9/5*-40+32#
#-40=-40#

#QED#

May 24, 2017

#C=-40#

Explanation:

#F=9/5C+32#

Let #F=C# so #C=9/5C+32#

Bring all #C# terms to one side

#-4/5C=32#

Divide both sides by #-4/5# to get #C# on its own

#C=-40#