"Question 8 In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: (i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis. (ii) If the temperature is 30°C, what is the temperature in Fahrenheit? (iii) If the temperature is 95°F, what is the temperature in Celsius? (iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius? (v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Class 9 - Math - Linear Equations in Two Variables Page 75"
Answers
Method to draw the graph of linear equation in two variables:
1. Let the linear equation in two variables be ax+by+c=0
Write the linear equation and Express Y in terms of x.
Y=(-ax+c)/b......(1)
2.Put put arbitrary value of x in equation 1 and find the corresponding values of y.
3.Form a table by writing the value’s of y to the corresponding values of x.
4.Draw the coordinates axes on graph paper and take a suitable scale to plot points from the table on graph paper.
5.Join the points and we get a straight line and produced it on both sides.
Hence, the straight line so obtained is the required graph of given linear equation.
It is enough to plot two points Corresponding to two solutions and join them by a line.
-----------------------------------------------------------------------------------------------------
Solution:
Given, linear equation in Fahrenheit and Celsius is
F=(9/5)C +32
5F-9C =160.........(1)
When C=0, then F= 160/5=32
When F= 0, then C = -160/9= -17.8
[Table and graph are on the attachment ]
ii)
If temperature is 30°C i.e C= 30°C
Then from equation 1 we get,
5F-9C=160
5F – 9×30=160
5F-270= 160
5F= 160+270
5F= 430
F=430/5=86°F
Temperature in Fahrenheit= 86°F
iii)
If temperature is 95° F , F= 95°F
Then from equation 1 ,
5F-9C=160
5×95 – 9C= 160
475-9C = 160
475-160= 9C
315= 9C
C= 315/9= 35°
Temperature in Celsius = 35°C
iv)
If the temperature is 0°C
Then, from eq 1
5F-9C=160
5F-9×0=160
5F=160
F=160/5=32
Temperature in Fahrenheit = 32°F
If the temperature is 0°F
Then, from eq 1
5F-9C=160
5×0-9×C=160
-9C=160
C=-160/9= -17.8°C
Temperature in Celsius= -17.8°C
v)
Yes, if we take both temperature are equal, i.e, C=F
Now ,from eq 1 we get
5F-9C=160
5F= 9C+160
5F= 9F+160
5F-9F=160
-4F= 160
F= -160/4= -40
Hence, F=C=-40°
=========================================================
Hope this will help you....
The Answer of your Question is ->
≥ (i) F = (9/5)C + 32
When C = 0 then F = 32
also, when C = -10 then F = 14
(ii) Putting the value of C = 30 in F = (9/5)C + 32, we get
F = (9/5)×30 + 32
⇒ F = 54 + 32
⇒ F = 86
(iii) Putting the value of F = 95 in F = (9/5)C + 32, we get
95 = (9/5)C + 32
⇒ (9/5)C = 95 - 32
⇒ C = 63 × 5/9
⇒ C = 35
(iv) Putting the value of F = 0 in F = (9/5)C + 32, we get
0 = (9/5)C + 32
⇒ (9/5)C = -32
⇒ C = -32 × 5/9
⇒ C = -160/9
Putting the value of C = 0 in F = (9/5)C + 32, we get
F = (9/5)× 0 + 32
⇒ F = 32
(v) Here, we have to find when F = C.
Therefore, Putting F = C in F = (9/5)C + 32, we get
F = (9/5)F + 32
⇒ F - 9/5 F = 32
⇒ -4/5 F = 32
⇒ F = -40
Therefore at -40, both Fahrenheit and Celsius numerically the same.