When Colin simplified the expressions below, he was surprised to find that they all gave the same solution! Try it for yourself.

$$3(x+6y) + 2(x-5y)$$$$4(2x-y) - 3(x-4y)$$$$-2(5x-y) + 3(5x+2y)$$

Here is a set of five expressions: $$(x + y) \quad (x + 2y) \quad (x - 2y) \quad (x + 4y) \quad (2x + 3y)$$

Choose any pair of expressions and add together multiples of each (like Colin did).

Warning... you will have to multiply the expressions by fractions in some cases.

Charlie's trial and improvement approach:

Charlie chose a value for $a$ and worked out the value of $b$ that gave $5x$.

He then kept adjusting the values of $a$ and $b$ until he also got $8y$:

$a$ |
$b$ | $a(x+2y) + b(2x+3y)$ |

$5$ |
$0$ | $5x+10y$ |

$4$ |
$\frac {1}{2}$ | $5x+9\frac {1}{2}y$ |

$3$ |
1 | $5x+9y$ |

$2$ |
$\frac {3}{2}$ | $5x+8\frac {1}{2}y$ |

$1$ |
2 | $5x+8y$ |

Alison's algebraic approach:

Alison multiplied out the brackets:$$\eqalign{a(x+2y)+b(2x+3y)&=5x+8y \\ \Rightarrow \begin{cases}ax+2bx &= 5x\\ 2ay+3by &= 8y \end{cases} \\ \Rightarrow \begin{cases} a+2b &= 5 \\ 2a+3b &= 8 \end{cases} } \\ \Rightarrow a=1 \quad \text{and} \quad b=2$$