# Can They Be Equal? Poster

## Problem

This poster is based on the problem Can They Be Equal?.

The poster is available as a PDF, or the image below can be clicked on to enlarge it.

## Student Solutions

Using whole numbers only: 3 by 6 and 4 by 4

Using fractions: Any two numbers $a$ and $b$ where $a\gt2$ and $b = \frac{2a}{a-2}$

**Why?**

__Geometrically__

There are only two ways to make a rectangle using 4 squares.

These make 4 by 4 or 3 by 6 rectangles.

This extends to fractions if the sides of the red rectangle aren't whole numbers, but any two numbers which multiply to 4.

__Algebraically__

Sides $a$ and $b$

Perimeter: $2a+2b$

Area: $ab$

$2a+2b = ab$

*Factorising to get whole number solutions:*

$ab - 2a-2b = 0$

$(a-2)(b-2) = \underbrace{ab-2a-2b}_0+4 \therefore (a-2)(b-2)=4$

(This fits with the geometry above because the sides of the red rectangle are $a-2$ and $b-2$)

Whole number solutions: $a-2$ and $b-2$ are $1$ and $3$ or $2$ and $2$

$\therefore a$ and $b$ are $3$ and $6$ or $4$ and $4$

Other solutions: Find any two numbers which multiply to give $4$ and add $2$ to each of them

*Rearranging to get infinitely many solutions*

$\begin{align}ab-2b &= 2a\\

\Rightarrow b(a-2)&=2a\\

\Rightarrow b&=\frac{2a}{a-2}\end{align}$

$a$ and $b$ need to be positive $\therefore a\gt2.$ Infinitely many solutions.