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: cost=2.85×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 | 8×60 | = | 5×92 | + | w |
calculate | 480 | = | 460 | + | w |
Subtract 460 from both sides | 20 | = | w |
Answer: 20 minutes