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# Always Two

$$ab+c=2\\ bc+a=2\\ca+b=2$$

You could start by subtracting one equation from another.

Alternatively, all of the equations are equal to $2$, so you could put two of the left hand sides equal to each other.

You should be able to find a common factor of $(a-c)$.

If you know that $(x-2)(x-3)=0$, then you can conclude that either $x-2=0$ or $x-3=0$. What are the two different cases for $(b-1)(a-c)=0$?

If $(b-1)(a-c)=0$, then we must have either $b=1$ or $a=c$.

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Age 14 to 18

Challenge Level

Three numbers, $a, b$ and $c$, have the property that when we multiply two of them together and then add the third one we always get the answer $2$.

Can you write down three equations that $a, b$ and $c$ satisfy?

Click on the button below to reveal the three equations that Claire wrote down.

$$ab+c=2\\ bc+a=2\\ca+b=2$$

Use two of your equations to show that: $$(b-1)(a-c)=0$$

You could start by subtracting one equation from another.

Alternatively, all of the equations are equal to $2$, so you could put two of the left hand sides equal to each other.

You should be able to find a common factor of $(a-c)$.

If we have $(b-1)(a-c)=0$, then what are the two cases that we must consider?

If you know that $(x-2)(x-3)=0$, then you can conclude that either $x-2=0$ or $x-3=0$. What are the two different cases for $(b-1)(a-c)=0$?

If $(b-1)(a-c)=0$, then we must have either $b=1$ or $a=c$.

Use one of your cases to reduce the original set of three equations to a set of two equations in two unknowns. Solve these to find one or more possible sets of solutions to the original problem. Repeat with the other case! There is more help available with this step in the getting started section.

**Extension**

Can you find all the possible values of $a, b$ and $c$ in these cases:

1. Three numbers, $a, b$ and $c$, have the property that when we multiply two of them together and then add the third one we always get the answer $6$.

2. Three numbers, $a, b$ and $c$, have the property that when we multiply two of them together and then **subtract** the third one we always get the answer $2$.

**Further extension**

Can you solve the general case where if you multiply two of the numbers together and then add the third you always get the value $k$?