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Lily from Jersey College for Girls found another possibility:
I found one possibilty for this question, and my answer became $3(x+2y) + 2(x+y)$, this would lead to $3x+6y + 2x+2y$, then the answer becoming $5x + 8y$.


Minjoon from Seoul Foreign British School found another one:
$a(x+y)+b(x-2y)$

$6x+6y-x+2y=5x+8y$
(so $a=6$ and $b=-1$)


Brian found a possibility using Alison's algebraic method:
$$a(x+2y)+b(2x+3y)=5x+8y\\\Rightarrow\begin{cases}ax+2bx=5x \\2ay+3by=8y\end{cases}\\
\Rightarrow\begin{cases} a+2b=5\\2a+3b=8\end{cases}\\
\Rightarrow a=1\hspace{4mm}\text{and}\hspace{4mm}b=2$$


Rishika from Nonsuch High School for Girls used her own algebraic method, similar to Alison's, to find all of the possibilities:
I used an algebraic method to find the multiples using the following method:
For example, $$a(x+y)+b(x+2y) = 5x+8y$$
{$a$ and $b$ represent the values of the multiples we need to find}

We can then form two equations if we substitute $x=0$ and $y=0$,
When $x=0$,
$$ay+2by = 8y$$
When $y=0$,
$$ax+bx = 5x$$
Then, we can cancel the like terms in each equation:

Equation 1: $a+2b = 8$

Equation 2: $a+b = 5$

 
Which we can solve simultaneously:

Equation 1 - Equation 2 (as '$a$' is present in both): $b = 3$

Substitute into Equation 2: $a = 2$
 
Therefore, $2 (x+y) +3 (x+2y) = 5x+8y$


I repeated this for all other possibilities.
Rishika made some mistakes, but these are the ones that Rishika found correctly:
$$\begin{align}2 (x+y) &+3 (x+2y) \hspace{4mm}\text{{from above}}\\
6 (x+y) &-1 (x-2y)\\
4 (x+y) &+1 (x+4y)\\
1 (x+2y) &+2 (2x+3y)\\
\end{align}$$


Rishika identified all of the possible pairs of expressions. Can you use Rikisha's method, or any other method, to find multiples of these pairs that add up to $5x+8y$?
$(x+y)$ and $(2x+3y)$
$ (x+2y)$ and $(x-2y)$
$ (x+2y)$ and $(x+4y)$
$ (x-2y)$ and $ (x+4y)$
$(x-2y) $ and $ (2x+3y)$
$(x+4y)$ and $ (2x+3y)$