Question 612782
Water boils at {{{100^o}}}C (degrees Celsius) which is {{{212^o}}}F (degrees Fahrenheit) and freezes at {{{0^o}}}C which is {{{32^o}}}F. Let x denote the temperature of an object in  F and y the temperature of the same object in  C. 
So we have two points (A and B) of the line (the linear function) that relates y and x.
{{{drawing(300,300,-50,200,-60,240,
grid(1),
locate(-48,238,x(degreesF)),locate(140,20,y(degreesC)),
line(190,-10,205,-10),
blue(circle(0,32,5)),blue(circle(100,212,5)),
locate(1,32,A(0,32)),locate(101,212,B(100,212)),
blue(line(-50,-58,200,392)),
locate(80,120,x=(9/5)*y+32)
)}}} or {{{drawing(300,300,-60,240,-50,200,
grid(1),
locate(-58,199,y(degreesC)),locate(170,15,x(degreesF)),
blue(circle(32,0,5)),blue(circle(212,100,5)),
locate(5,15,A(32,0)),locate(130,115,B(212,100)),
blue(line(-58,-50,392,200)),
locate(65,176,y=(5/9)*(x-32))
)}}}
With the coordinates of two points, you can determine the slope and the equation of the line.
 
(a) Express x in terms of y. 
x(y)= F 
In the graph on the left, x increases by {{{212^o}}}F-{{{32^o}}}F={{{180^o}}}F (the vertical rise from A to B),
while y increases by {{{100^o}}}C-{{{0^o}}}C={{{100^o}}}C (the horizontal run from A to B).
So the slope is {{{(212-32)/(100-0)=180/100=9/5}}}
With that slope, and the y-intercept {{{32^oF}}} from point A, we can write the slope-intercept form of the equation as
{{{highlight(x(y)=(9/5)*y+32)}}}F
 
(b) Express y in terms of x. 
y(x)= C
You start from scratch calculating the slope of the line, and then using that slope and point A to get the point slope form of the equation.
However, the easiest way to get the needed equation is to solve for y in the equation found in part (a).
{{{x=(9/5)*y+32}}} --> {{{x-32=(9/5)*y}}} --> {{{(5/9)(x-32)=(5/9)(9/5)*y}}} -->  {{{(5/9)(x-32)=y}}}
I would stop at {{{highlight(y(x)=(5/9)*(x-32))}}}C, because that is the equation I would use to do mental math conversions.
Transforming that into the slope-intercept form would give you a less user-friendly equation:
{{{y(x)=(5/9)*(x-32)}}} --> {{{x(y)=(5/9)*y-(5/9)*32}}} --> {{{x(y)=(5/9)*y-160/9}}} 
 
(c) Use (b) to convert  F into Celsius. 
If I wanted to convert {{{105^o}}}F into degrees Celsius to tell a very ill tourist his/her body temperature in familiar units, I would use the equation from part (b), to get
{{{y(105)=(5/9)*(105-32)=(5/9)*73=40.6}}}C (rounded from 40.55555....)
 
(d) Use (a) to convert your answer to (c) back into Fahrenheit. 
{{{x(40.6)=(9/5)*40.6+32)=105.1}}}F (rounded from 105.08)
Yes, rounding errors caused the final answer to be a tiny bit different from the {{{105^o}}}F I started with.