Exercise 23 Page 46

Practice makes perfect

a In being asked to solve the given formula for K, we are actually being asked to isolate the variable K.

F=9/5(K-273.15)+32

LHS-32=RHS-32

F-32=9/5(K-273.15)

LHS * 5=RHS* 5

5F-160=9(K-273.15)

Distribute 9

5F-160=9K-2458.35

LHS+2458.35=RHS+2458.35

5F+2298.35=9K

.LHS /9.=.RHS /9.

5F+2298.35/9=K

Rearrange equation

K=5F+2298.35/9

b We can use our formula from the previous exercise to convert 180^(∘)F to kelvin K.

K=5F+2298.35/9

F= 180

K=5 * 180+2298.35/9

Use a calculator

K=355.3722222

Round to 2 decimal place(s)

K ≈ 355.37

180^(∘)F is equal to 355.37K.

1 Solving Simple Equations p.44

2 Solving Multi-Step Equations p.44

3 Solving Equations with Variables on Both Sides p.45

4 Solving Absolute Value Equations p.45

5 Rewriting Equations and Formulas p.46

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