# When solving a rational equation, why is it necessary to perform a check?

Jun 11, 2015

It is necessary to perform a check because in the process of multiplying through you can introduce spurious solutions.

#### Explanation:

Consider the example:

$\frac{x + 3}{{x}^{2} - 3 x + 2} = \frac{x + 2}{{x}^{2} - 4 x + 3}$

We could choose to "cross multiply" the equation to get:

$\left(x + 3\right) \left({x}^{2} - 4 x + 3\right) = \left(x + 2\right) \left({x}^{2} - 3 x + 2\right)$

That is:

${x}^{3} - {x}^{2} - 9 x + 9 = {x}^{3} - {x}^{2} - 4 x + 4$

Subtract ${x}^{3} - {x}^{2}$ from both sides to get:

$- 9 x + 9 = - 4 x + 4$

Add $4 x - 4$ to both sides to get:

$- 5 x + 5 = 0$

Divide both sides by $5$ to get

$- x + 1 = 0$

Hence $x = 1$

But try putting $x = 1$ in the original equation and you will find both denominators are zero.

What went wrong here is that both $\left({x}^{2} - 3 x + 2\right)$ and $\left({x}^{2} - 4 x + 3\right)$ are divisible by $\left(x - 1\right)$, so cross multiplying by them included the effect of multiplying both sides by ${\left(x - 1\right)}^{2}$ - not only clearing $\left(x - 1\right)$ from the denominator, but adding an extra factor of $\left(x - 1\right)$ on both sides of the equation.