The Simple Life
The answer is $5x+8y$... What was the question?
Age
11 to 14
Challenge level
Problem
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 $8y$:
$a$ |
$b$ | $a(x+2y) + b(2x+3y)$ |
$5$ |
$0$ | $5x+10y$ |
$4$ |
$\frac {1}{2}$ | $5x+9\frac {1}{2}y$ |
$3$ |
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:$$\eqalign{a(x+2y)+b(2x+3y)&=5x+8y \\ \Rightarrow \begin{cases}ax+2bx &= 5x\\ 2ay+3by &= 8y \end{cases} \\ \Rightarrow \begin{cases} a+2b &= 5 \\ 2a+3b &= 8 \end{cases} } \\ \Rightarrow a=1 \quad \text{and} \quad b=2$$
With thanks to Colin Foster who introduced us to this problem.
Getting Started
For each of the pairs below, there are whole number multiples that combine to give $5x + 8y$:
$$(x + y) \quad (x + 2y)$$$$(x + y) \quad (x + 4y)$$$$(x + 2y) \quad (2x + 3y)$$$$(2x + 3y) \quad (x + y)$$$$(x - 2y) \quad (x + 4y) $$$$(x + y) \quad (x - 2y)$$$$(x + 2y) \quad (x + 4y)$$
Student Solutions
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 2: $a+b = 5$
Equation 1 - Equation 2 (as '$a$' is present in both): $b = 3$
Substitute into Equation 2: $a = 2$
Therefore, $2 (x+y) +3 (x+2y) = 5x+8y$Substitute into Equation 2: $a = 2$
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)$
Teachers' Resources
The suggestions in these notes are adapted from Colin Foster's article, The Simple Life
Why do this problem?
Expanding brackets and collecting like terms to simplify expressions are perhaps not the most exciting of mathematical topics, but fluency in these skills is needed in order to solve equations and engage with more stimulating mathematical problem solving.One way to avoid the tedium of lots of repetitive practice is to embed practice in a bigger problem which students are trying to solve. This idea is explored in Colin Foster's article, Mathematical Etudes, and this problem is an example of a mathematical etude.
Possible approach
"I would like you to expand and simplify these four expressions. Which one is the odd one out?"- $(3x + 4y) + 2(x + 2y)$
- $4(2x + 5y) - 3(x + 4y)$
- $3(2x + 3y) - (x - y)$
- $3(x + 3y) + (2x - y)$
When students multiply out the brackets and collect like terms they should find that (1), (2) and (4) all come to $5x + 8y$, so (3) is the odd one out. Students might get $5x + 8y$ instead of $5x+10y$ for (c) if they subtract $-y$ incorrectly.
"Was anything tricky about these?"
"Did you make any mistakes? Or nearly make any mistakes?"
"What things do you have to be careful of when simplifying expressions like this?"
"Do you have any advice for someone doing questions like this?"
"Today, your challenge is to combine pairs of expressions to get $5x + 8y.$"
"The only expressions that you are allowed to use are: $$(x + y) \quad (x + 2y) \quad (x - 2y) \quad (x + 4y) \quad (2x + 3y)$$ You can pick any two of these expressions and add or subtract multiples of each.
"For example, we could choose the brackets $(x + 2y)$ and $(x + 4y)$" $$\square(x + 2y) ± \square(x + 4y) = 5x+8y$$ "Now we need to find numbers to go in the boxes."
Ask students to suggest some numbers and then work through the expanding and simplifying process together on the board. It is very unlikely that the answer will come to 5x + 8y.
"Can you choose a different pair of numbers so that we get $5x + 8y$?"
Invite the class to work on the problem in pairs. Once they find the answer, they can choose different pairs of expressions and repeat the task.
Seven pairs of expressions use whole numbers, and three pairs require fractions - see Getting Started for a list.
While they are working, circulate and look out for students using efficient strategies, like the two listed in the problem. After a while, bring the class together and invite students to talk about their methods. If no-one has approached the problem using Charlie's or Alison's methods, share them with the class.
Then give everyone a little longer to work on the problem, perhaps encouraging them to try out both methods. At the end of the lesson, you could invite students to discuss the pros and cons of each method.
You could conclude the lesson with a plenary in which the students talk about their answers and how they got them.