Equations can be derived :
• when the value of one item varies with another;
• there is a rate (often seen with the word per) that multiplies the item;
• a starting value, which is added (or subtracted).
If the change is linear (a straight line), the data items can be laid out as an equation:
result = rate x item + starting value
which corresponds to
`y=mx+c`
A taxi firm charges £2.85 per kilometre, plus a £2 hire charge. If my taxi fare was £17.96, what distance did I travel?
The equation is: `text(cost) = 2.85 xx text(distance) + 2`
| Rewrite as | `c` | `=` | `2.85d` | `+` | `2` |
| Substituting | `17.96` | `=` | `2.85d` | `+` | `2` |
| Subtract 2 from both sides | `15.96` | `=` | `2.85d` | `` | `` |
| Divide both sides by 2.85 | `5.6` | `=` | `d` | `` | `` |
Answer: 5.6 kilometers
A company produces a complicated part for a car. The machine takes time to set up before it can produce the parts: after it has been set up, it produces 1 part every 5 minutes.
If the machine produces 92 parts in an 8-hour shift, how long is the set-up time?
| Create an equation | shift | = | rate x parts | + | warmup |
| replace with letters | `s` | `=` | `rp` | `+` | `w` |
| Substitute, using minutes | `8xx60` | `=` | `5xx92` | `+` | `w` |
| calculate | `480` | `=` | `460` | `+` | `w` |
| Subtract 460 from both sides | `20` | `=` | `` | `` | `w` |
Answer: 20 minutes