Question 263671
A solution provided by another tutor is correct as far as it goes. But it is missing a critical last step!<br>
The solution provided multiplies both sides of the equation by (z+2)(z-2). Multiplying both sides of an equation by an expression that might end up being zero (and (z+2)(z-2) might be zero depending on the value fo z) may introduce what are called extraneous solutions. Extraneous solutions are solutions that work in the equation after this multiplication <i>but do not work in the original equation!</i> So we must check our solution(s)!<br>
The other tutor's solutions are 2 and -2. Let's check them:
{{{5z/(z-2) - 10/(z+2) = 40/(z^2 -4)}}}
Checking z = 2:
{{{5(2)/((2)-2) - 10/((2)+2) = 40/((2)^2 -4)}}}
Simplifying we get:
{{{10/0 - 10/4 = 40/0}}}
As we can see, we get some zero denominators!! So z = 2 is an extraneous solution and must be rejected. (If only one denominator was zero we would still have to reject this solution.)<br>
Checking z = -2:
{{{5(-2)/((-2)-2) - 10/((-2)+2) = 40/((-2)^2 -4)}}}
Simplifying we get:
{{{(-10)/(-4) - 10/0 = 40/0}}}
Again we get some zero denominators!! So z = -2 is also an extraneous solution and must be rejected. (If only one denominator was zero we would still have to reject this solution.)<br>
We have rejected both solutions! This means that your original equation <i>has no solution!</i>