# Teach Yourself - Algebra A complete introduction E6.2 Q13

Linear, quadratic, module, parametric equations

### Teach Yourself - Algebra A complete introduction E6.2 Q13

Can someone explain how to get to the answer?

Thanks!
Attachments
7526B5A3-BD20-463E-812A-FA8E630A61CE.jpeg (410.24 KiB) Viewed 439 times
Guest

### Re: Teach Yourself - Algebra A complete introduction E6.2 Q1

You're given the conversion formula

$$x ^\circ F = \left( \frac{9x}{5}+32\right)^\circ C$$

If you substitute $$x=86$$ into that formula you get

$$86 ^\circ F = \left( \frac{9 \cdot 86}{5}+32\right)^\circ C = 186.8^\circ C$$

This where you go, "Bullshit, 86F is a pleasant sunny day. 186C is the temperature for baking a cake." That's because your book made a mistake and got the conversion reversed. They should have written

$$x ^\circ C = \left( \frac{9x}{5}+32\right)^\circ F$$

In this case, $$x$$ represents degrees Celsius which is what you want to solve for, and you are given that $$\frac{9x}{5}+32 = 86$$. Subtract 32 for $$\frac{9x}{5} = 54$$ and multiply by 5/9 for $$x=54*5/9 = 30$$. 86F = 30C is much more plausible and is in fact the correct conversion.

phw

Posts: 11
Joined: Sun May 06, 2018 1:15 am
Reputation: 7