Processing Math: Done

From Förberedande kurs i matematik 1

Start by rewriting the terms on the left-hand side as one term having a common denominator

1x−1−1x+1=1x−1x+1x+1−1x+1x−1x−1=x+1(x−1)(x+1)−x−1(x−1)(x+1)=(x−1)(x+1)(x+1)−(x−1)=2(x−1)(x+1).

If we also write 3x−3=3(x−1), the equation can be rewritten as

2(x−1)(x+1)x2+21=6x−13(x−1).

Because x=1 cannot be a solution to the equation, the factor x−1 can be removed from the denominator of both sides (i.e. actually, we multiply both sides by x−1 and then eliminate it)

2x+1x2+21=36x−1.

Then, both sides are multiplied by 3 and x+1, so that we get an equation without any denominators

6x2+21=(6x−1)(x+1).

Expanding both sides

6x2+3=6x2+5x−1.

The x² terms cancel each other out and we obtain a first-order equation,

3=5x−1

which has the solution

x=54.

We check whether we have calculated correctly by substituting x=45 into the original equation,

LHSRHS=154−1−154+1542+21=1−51−1592516+21=−5−95225162+25=−9505057=−957=−319=354−3654−1=524−55512−515=51(24−5)51(12−15)=19−3=−319.