# At what temperature is the Fahrenheit scale reading equal to (a) twice (b) half of the Celsius?

We know that,

TC = (TF – 32) x $frac{5}{9}$ (equ.1)

(a) Twice the Celsius

TF = 2TC

From the equ.1, we can substitute for TF

TC = (2TC – 32) x $frac{5}{9}$

9TC = 10TC – 160

TC = 160°C

(b) Half of the Celsius

TF = $frac{T_{C}}{2}$

From the equ.1, we can substitute for TF

TC = ($frac{T_{C}}{2}$ – 32) x $frac{5}{9}$

9TC = $frac{5}{2}T_{C}$ – 160

TC = -24.6°C