Originally Posted by

**mr fantastic**

*Unique solution:*

x + by = -1 .... (1)

ax + 2y = 5 .... (2)

Solve using the elimination method:

Multiply equation (1) by a, subtract it from equation (2) and then solve for y:

$\displaystyle (2 - ab)y = 5 + a \Rightarrow y = \frac{5 + a}{2 - ab}$.

Note that this solution is OK *provided the denominator is NOT equal to zero*, that is, $\displaystyle 2 - ab \neq 0 \Rightarrow ab \ne 2$.

Now sub $\displaystyle y = \frac{5 + a}{2 - ab}$ into either equation and solve for x. Using equation (1):

$\displaystyle x + by = -1 \Rightarrow x = -1 - by = -1 - b \left( \frac{5 + a}{2 - ab} \right)$ $\displaystyle = \frac{-(2 - ab)}{2 - ab} - \frac{b(5 + a)}{2 - ab} = \frac{-2 + ab - 5b - ab}{2 - ab} = \frac{-2 - 5b}{2 - ab}$.

So there's a unique solution, given by the above, when $\displaystyle ab \neq 2$.

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No unique solution: $\displaystyle ab = 2$.

When there's no unique solution, that means that either there is an infinite number of solutions or no solution at all. To test which of these happens when ab = 2, you might do something like the following:

$\displaystyle ab = 2 \Rightarrow b = \frac{2}{a}$. Substitute this into you equations and do some re-arranging:

Equation (1): $\displaystyle x + by = -1 \Rightarrow x + \frac{2}{a} y = -1 \Rightarrow ax + 2y = -a$.

Equation (2): $\displaystyle ax + 2y = 5$.

Now compare equations (1) and (2). You can see that when $\displaystyle -a = 5 \Rightarrow a = -5$ the equations are the same. One equation and two unknowns => infinite number of solutions.

So when ab = 2 AND a = -5, there an infinite number of solutions.

It follows that there will be no solution at all when ab = 2 AND $\displaystyle a \neq -5$.

When a = -5 and ab = 2, $\displaystyle b = -\frac{2}{5}$. So the equations boil down to -5x + 2y = 5. The form of the infinite number of solutions can be expressed in several ways. To see where the books answer came from, note that if you let y = t, where t can be any real number:

$\displaystyle -5x + 2t = 5 \Rightarrow -5x = 5 - 2t \Rightarrow 5x = 2t - 5 \Rightarrow x = \frac{2}{5}t - 1$.