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Age 11 to 14 Challenge Level:

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).

Can you find a way to get an answer of $5x+8y$ in each case?

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

If you're struggling to get started... take a look below to see how Charlie and Alison thought about the problem when combining multiples of $(x+2y)$ and $(2x+3y)$.

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 $8y$:

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

$0$ $5x+10y$

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

1 $5x+9y$

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

2 $5x+8y$

Alison's algebraic approach:

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

With thanks to Colin Foster who introduced us to this problem.