### Video Transcript

Show that π’ over π£ minus one
squared plus π’ over π£ minus one can be simplified to π’π£ over π£ minus one
squared, with π£ is not equal to one.

This may look nasty. But, really, itβs asking us to add
two fractions together. It just so happens that these
fractions contain algebra. Letβs think about how we would
normally add fractions.

Weβd first find a common
denominator and then we create equivalent fractions before finally adding the
numerators. In fact, we need to follow roughly
the same process here. Notice how both of the dominators
contain π£ minus one. We can multiply the denominator of
the second fraction by π£ minus one to get π£ minus one squared. Thatβs our common denominator.

If we do that, we must also
multiply the numerator of the second fraction by π£ minus one. That gives us π’ over π£ minus one
squared β that first fraction hasnβt changed β plus π’ multiplied by π£ minus one
all over π£ minus one squared.

Notice how the denominator in our
answer now looks like the denominator required of us in the question. A sensible next step is to expand
these brackets. Remember we can expand a single
bracket by multiplying the term on the outside by everything on the inside: π’
multiplied by π£ is π’π£ and π’ multiplied by negative one is negative π’. The second fraction, therefore,
becomes π’π£ minus π’ all over π£ minus one squared.

Now that the denominators are the
same, we can add the numerators. That gives us π’ plus π’π£ minus π’
all over π£ minus one squared. We should then try to simplify by
collecting like terms. π’ minus π’ is simply zero.

Our expression simplifies to π’π£
over π£ minus one all squared as required.