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

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