A candle is 7 in tall after burning for 1 h. The same candle is 5.5 in tall after burning for 4 h. How tall will the candle be after burning for 6 h?

graph{x = (15 - y)/2 [-1, 10, -4.56, 16]}

We know that the length of the candle changed from 7 inches to 5.5 inches between time 1 hour and 4 hours. The rate of change can be calculated in inches per hour.

#m = (Delta h)/(Delta t) = (h_f - h_i)/(t_f - t_i)#
#m = (5.5-7)/(4 - 1) = 1.5/3 = -1/2#

The height at time 0 can be calculated by plugging in the slope #m# into our line equation:
#y = mx + b#
We know #y# at time 4 hours is 7 inches.
#7 = -1/2 * 1 + b#
#7 = -1/2 + b#
#7.5 = b#

Plugging both of those values into the equation and using the variables #l# for the length of the candle and #t# for time we get:
#l = -1/2 * t + 7.5#

And we can calculate that at time #t=6# the length of the candle will be:
#l = -1/2 * 6 + 7.5#
#l = -3 + 7.5#
#l = 4.5#

Here's a video of a candle burning over a couple of hours to inspire you:
Here's a video

The time to reduce to the 7 inches tall is of importance. Unfortunately we do not have the starting height.

#color(blue)("Determine the ratio of the burning rate")#

#("distance B to C")/("time B to C")->(7-5.5)/(4-1)color(magenta)(=1.5/3" as a ratio")#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the measurement B to D "-> x)#

#("distance B to D")/("time B to D")-> x/(6-1) = x/5#

This will have the same ratio as our previously calculated burning rate giving:

#1.5/3-=x/5# where #-=# means equivalent to.

Matching denominators: to change 3 to 5 we multiply by #5/3#

So if we multiply the denominator by #5/3# we have to do the same to the numerator.

#color(green)(1.5/3color(red)(xx1) " "->" "1.5/3color(red)(xx(color(white)(.)5/3color(white)(.))/(5/3))" "->" "2.5/5=("distance B to D")/("time B to D")#

#color(magenta)(x/5=2.5/5=> x=2.5)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine distance D to E")#

#h=DE" "=" "BE-x" " =" " 7-2.5 = 4.5 #