Hello Stroodle

This question is all about making sure you never divide by something that might be zero. So, continuing from here: Originally Posted by

**Prove It** $\displaystyle ax + y + z = 1$

$\displaystyle (ab - 1)y + (a - 1)z = a - 1$

$\displaystyle (b - 1)y = 0$

If $\displaystyle b \ne 1$, we can divide both sides of $\displaystyle (b-1)y=0$ ...(4)

by $\displaystyle (b-1)$ to get:$\displaystyle y = 0$

Substitute into (2):$\displaystyle (a-1)z = a-1$ ...(5)

So if $\displaystyle a\ne 1$ we can now divide both sides by $\displaystyle (a-1)$ to get:$\displaystyle z=1$

Substitute these values of $\displaystyle y$ and $\displaystyle z$ into (1) to get:$\displaystyle ax+0+1=1$

$\displaystyle \Rightarrow ax = 0$ ...(6)

Again, we must ensure that we don't divide by zero. So, if $\displaystyle a \ne 0$, we can divide both sides by $\displaystyle a$ and get:$\displaystyle x=0$

Therefore, provided $\displaystyle b \ne 1$ and $\displaystyle a \ne 1$ and $\displaystyle a \ne 0$, the solution is:$\displaystyle (0,0,1)$

Now we need to re-trace our steps, and see what happens if one or more of these conditions is not satisfied.

Starting with the last condition first, if $\displaystyle a = 0$, then equation (6) is satisfied by any value of $\displaystyle x$, say $\displaystyle x = \lambda$. So, if $\displaystyle b \ne 1$ and $\displaystyle a = 0$ we get the solution:$\displaystyle (\lambda, 0,1)$, for any value of $\displaystyle \lambda$.

Next, if $\displaystyle b \ne 1$ and $\displaystyle a = 1$, then (5) is satisfied by any value of $\displaystyle z$, say $\displaystyle z = \mu$. Substituting this value into (1) (with $\displaystyle a = 1$ and $\displaystyle y = 0$) we get:$\displaystyle x+\mu = 1$

$\displaystyle \Rightarrow x = 1-\mu$

So the solution for $\displaystyle b \ne 1$ and $\displaystyle a = 1$ is:$\displaystyle (1-\mu,0,\mu)$, for any value of $\displaystyle \mu$.

Now you need to go back and see what happens if $\displaystyle b = 1$, where equation (4) is then satisfied by any value of $\displaystyle y$.

Grandad