It is given that p=q+r2q−5p = \dfrac{q + r^2}{q - 5}.

  1. (a)

    Find pp when q=15q = \dfrac{1}{5} and r=−6r = -6.

    Answer p = ...

  2. (b)

    Express qq in terms of pp and rr.

    Answer ...

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Worked solution

9(a)

1. Substitute $q=\dfrac15$ and $r=-6$ into the formula and evaluate.

p=q+r2q−5p=\dfrac{q+r^2}{q-5}

=15+(−6)215−5=\dfrac{\dfrac15+(-6)^2}{\dfrac15-5}

=15+3615−5=\dfrac{\dfrac15+36}{\dfrac15-5}

=1815−245=\dfrac{\dfrac{181}{5}}{-\dfrac{24}{5}}

=−18124=-\dfrac{181}{24}

Answer

p=−18124p=-\dfrac{181}{24}

9(b)

1. Clear the denominator by multiplying both sides by $q-5$.

p=q+r2q−5p=\dfrac{q+r^2}{q-5}

p(q−5)=q+r2p(q-5)=q+r^2

2. Collect the terms involving $q$ and factorise to isolate $q$.

pq−5p=q+r2pq-5p=q+r^2

pq−q=r2+5ppq-q=r^2+5p

q(p−1)=r2+5pq(p-1)=r^2+5p

q=r2+5pp−1q=\dfrac{r^2+5p}{p-1}

Answer

q=r2+5pp−1q=\dfrac{r^2+5p}{p-1}

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