Simultaneous equations involve two sets of variables. Deriving a simultaneous equation from a text involves:
• determining the two variables involved;
• identifying the multiples associated with each of the two variables;
• identifying the sum of each of the multiple + variable pairs.
At a garden centre, four shrubs and two trees cost £56. Five shrubs and one tree cost £52.
How much would an individual shrub cost?
| Let shrubs = `s` and trees = `t`: | `4s + 2t` | `=` | `56` |
| and | `5s + t` | `=` | `52` |
| Rearrange the second equation | `t` | `=` | `52 - 5s` |
| Substitute `t` into the 1st equation: | `4s+ 2(52- 5s)` | `=` | `56` |
| Expand the bracket: | `4s+ 104- 10s` | `=` | `56` |
| Add the s together | `104- 6s` | `=` | `56` |
| Add 6s to both sides | `104` | `=` | `56+ 6s` |
| Subtract 56 from both sides | `48` | `=` | `6s` |
| Divide both sides by 6 | `8` | `=` | `s` |
| Substitute into `5s+t=52` | `5(8) + t=52` | ||
| `t=12` | |||
| Check into `4s+2t=56` | `4(8) + 2(12)` | `=` | `56` ✔ |
Answer: shrub = £8
Two families went to the same restaurant. The Khans had 3 pizzas and one pasta; the Smiths had 2 pizzas and 2 pastas. The bill for the Khans was £35.00, which was £1.50 more than the bill for the Roberstons.
How much was a pizza at the restaurant?
| Let `x` = pizza and `y` = pasta | |||
| Robertsons: | `3x + y` | `=` | `35.00 - 1.50` |
| Khans: | `2x + 2y` | `=` | `35.00` |
| Rearrange 1st equation: | `y` | `=` | `33.5 - 3x` |
| Second equation: | `2x+ 2y` | `=` | `35` |
| Substitute into 2nd equation | `2x+ 2(33.5- 3x)` | `=` | `35` |
| Expand the brackets | `2x+ 67- 6x` | `=` | `35` |
| Add `4x` to both sides | `67` | `=` | `35+ 4x` |
| Subtract 35 from both sides | `32` | `=` | `4x` |
| Divide both sides by 4 | `8` | `=` | `x` |
| Substitute `x=8` into 2nd equation: | `2(8) + 2y=35` | ||
| `y=9.5` | |||
| Check: | `3(8) + 2(9.5)` | `=` | `33.5` ✔ |
Answer: £8.00