Results 1 to 3 of 3

Thread: Graph Involving Linear Equation on degrees

  1. #1
    Jan 2010

    Graph Involving Linear Equation on degrees

    Ok here is the Problem.

    Weather Temperature. The National Oceanic and Atmospheric ADministration (NO) has an online cconversion chart that relates degrees Fahrenheit to degrees Celsius. 77deg F is equevalent to 25deg C and 68deg F is equivalent to 20deg C. Assuming relationship is linear, write the equation relating degrees Celsius C to degrees Fahrenheit F. What temperature is the same in both degress Celsius and degress Fahrenheit?

    Here steps I did:

    77deg F = 25deg C
    68deg F = 20deg C

    Let x = Fahrenheit and y = Celsius
    Linear Equation: y=mx + b or C = mF + b

    Two Points of this Equation is (77,25), (68,20)

    Solving the equation:

    Calculate the change of rate:
    $\displaystyle \frac {Y2 + Y1} {X2-X1}$

    $\displaystyle m= \frac {25-20} {68-77} = \frac {-5} {-9} $

    $\displaystyle m= \frac {-5} {-9} $

    Subtitute the slope into the linear equation.

    $\displaystyle m= \frac {-5} {-9} x + b $

    finding b:

    I use either point to satisfy the equation. I use (77,25)
    Let x = 77
    C = \frac {-5}{-9}F + b $
    25= \frac {-5} {-9} (77) + b $
    25= \frac {385} {9} + b $
    b = \frac {160} {9}$

    $\displaystyle Equation is : C = \frac {-5} {-9}F + \frac {160} {9} $

    I don't know if i'm this right. I still haven't answer the other question: What temperature are both the same degrees.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Feb 2010
    you were correct throughout, but you made the slope negative in the denominator and numerator, making it positive . using what you said at first for the variables you would get the points (77,25) and (68,20).

    using slope formula you get:

    giving you a slope of $\displaystyle \frac{5}{9}$

    next i would use point slope form at first then rearrange,so..


    using the point (68,20)




    now to find when they are the do this set y=x






    so at -40 degrees they are the same
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Quacky's Avatar
    Nov 2009
    Windsor, South-East England
    Would you not just use the $\displaystyle y-y1=m(x-x1)$ formula?
    You have a point (77,25) so substituting that into the equation:
    $\displaystyle y-25=m(x-77)$

    And using the rule for calculating a gradient:
    $\displaystyle m=\frac{y2-y1}{x2-x1}$
    $\displaystyle m=\frac{25-20}{77-68}$
    $\displaystyle =\frac{5}{9}$

    $\displaystyle y-25=\frac{5}{9}(x-77)$
    $\displaystyle 9C-225=5(F-77)$
    $\displaystyle 9C-225=5F-385$
    $\displaystyle 9C=5F-160$
    Checking this by substituting the values back into it suggests that it is correct.

    For the second part, you are trying to find when C=F. So just substitute this into the equation:
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Question on quadratic and linear equation/graph
    Posted in the Statistics Forum
    Replies: 3
    Last Post: Sep 29th 2011, 06:22 AM
  2. Graph the linear equation with two variables?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Oct 10th 2010, 07:41 PM
  3. [SOLVED] Linear equation for log graph
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Mar 9th 2010, 04:34 AM
  4. Replies: 2
    Last Post: Apr 6th 2009, 09:19 AM
  5. Graph Linear Equation X+Y=5
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 25th 2007, 06:40 PM

Search Tags

/mathhelpforum @mathhelpforum