Introduction to Linear Polynomials | Exercise 2.5

Question 3

Consider the relationship between temperature measured in degrees Celsius (∘C^\circ\text{C}) and degrees Fahrenheit (∘F^\circ\text{F}), which is given by ∘C=a∘F+b^\circ\text{C} = a^\circ\text{F} + b. Find aa and bb, given that ice melts at 00 degrees Celsius and 3232 degrees Fahrenheit, and water boils at 100100 degrees Celsius and 212212 degrees Fahrenheit. (Hint: When ∘C=0^\circ\text{C} = 0, ∘F=32^\circ\text{F} = 32 and when ∘C=100^\circ\text{C} = 100, ∘F=212^\circ\text{F} = 212. Use this information to find aa and bb, and thus, the linear relationship between ∘C^\circ\text{C} and ∘F^\circ\text{F}.)

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Solution
Understand the Question
  • The temperature conversion formula is given as a linear equation: ∘C=a∘F+b^\circ\text{C} = a^\circ\text{F} + b.
  • We are given two data points (∘F,∘C)(^\circ\text{F}, ^\circ\text{C}):
    • Ice melts: (32,0)(32, 0)
    • Water boils: (212,100)(212, 100)
  • Substituting these two points into the equation gives a system of two linear equations in variables aa and bb, which can be solved simultaneously to find their values.

Step 1 · Set up Equations

Substitute the melting point of ice (∘C=0,∘F=32^\circ\text{C} = 0, ^\circ\text{F} = 32) into ∘C=a∘F+b^\circ\text{C} = a^\circ\text{F} + b

0=a(32)+b0=32a+b…(1)\begin{aligned} 0 &= a(32) + b \\ 0 &= 32a + b \quad \dots (1) \end{aligned}

Substitute the boiling point of water (∘C=100,∘F=212^\circ\text{C} = 100, ^\circ\text{F} = 212)

100=a(212)+b100=212a+b…(2)\begin{aligned} 100 &= a(212) + b \\ 100 &= 212a + b \quad \dots (2) \end{aligned}

Step 2 · Solve for aa

Subtract equation (1)(1) from equation (2)(2)

(212a+b)−(32a+b)=100−0212a−32a=100180a=100a=100180a=1018a=59\begin{aligned} (212a + b) - (32a + b) &= 100 - 0 \\ 212a - 32a &= 100 \\ 180a &= 100 \\[0.6em] a &= \dfrac{100}{180} \\[0.6em] a &= \dfrac{10}{18} \\[0.6em] a &= \dfrac{5}{9} \end{aligned}

Step 3 · Solve for bb

Substitute a=59a = \dfrac{5}{9} into equation (1)(1)

0=32(59)+b0=1609+bb=−1609\begin{aligned} 0 &= 32\left(\dfrac{5}{9}\right) + b \\[0.6em] 0 &= \dfrac{160}{9} + b \\[0.6em] b &= -\dfrac{160}{9} \end{aligned}

Step 4 · State the Linear Relationship

Substitute a=59a = \dfrac{5}{9} and b=−1609b = -\dfrac{160}{9} into ∘C=a∘F+b^\circ\text{C} = a^\circ\text{F} + b ∘C=59∘F−1609^\circ\text{C} = \dfrac{5}{9}^\circ\text{F} - \dfrac{160}{9}

Factoring out 59\dfrac{5}{9}

∘C=59(∘F−1605)=59(∘F−32)\begin{aligned} ^\circ\text{C} &= \dfrac{5}{9}\left(^\circ\text{F} - \dfrac{160}{5}\right) \\[0.6em] &= \dfrac{5}{9}(^\circ\text{F} - 32) \end{aligned}
Answer

a=59,b=−1609a = \dfrac{5}{9}, \quad b = -\dfrac{160}{9}

The linear relationship is ∘C=59(∘F−32)^\circ\text{C} = \dfrac{5}{9}(^\circ\text{F} - 32).

Common Mistakes
  • Swapping Celsius and Fahrenheit: Accidentally substituting ∘C^\circ\text{C} in place of ∘F^\circ\text{F} (e.g. writing 32=a(0)+b32 = a(0) + b instead of 0=a(32)+b0 = a(32) + b).
  • Sign Error when Isolating bb: Forgetting to change the sign when transposing 1609\dfrac{160}{9} to the other side: b=−1609b = -\dfrac{160}{9}, not +1609+\dfrac{160}{9}.

More questions in Exercise 2.5

Q1

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill yy depends on the number of modules accessed, xx, according to the relation y=ax+by = ax + b, find the values of aa and bb.

Q2

A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill yy depends on the hours of the use of the badminton court, xx, according to the relation y=ax+by = ax + b, find the values of aa and bb.

Q3

Consider the relationship between temperature measured in degrees Celsius (∘C^\circ\text{C}) and degrees Fahrenheit (∘F^\circ\text{F}), which is given by ∘C=a∘F+b^\circ\text{C} = a^\circ\text{F} + b. Find aa and bb, given that ice melts at 00 degrees Celsius and 3232 degrees Fahrenheit, and water boils at 100100 degrees Celsius and 212212 degrees Fahrenheit. (Hint: When ∘C=0^\circ\text{C} = 0, ∘F=32^\circ\text{F} = 32 and when ∘C=100^\circ\text{C} = 100, ∘F=212^\circ\text{F} = 212. Use this information to find aa and bb, and thus, the linear relationship between ∘C^\circ\text{C} and ∘F^\circ\text{F}.)

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