As an alternative to the solution above, working with the two equations you have been given, use substitution to solve.

`therefore p+q=11` becomes `p=11-q` when rearranged which now substitute into the other equation:

`therefore pq=18` becomes `(11-q) times q = 18`

`therefore 11q -q^2 = 18`

`therefore 0 = q^2-11q+18` (Note the equation has been rearranged)

Solve by factorizing using the first term's factors (q x q) and the third term's factors which will render a middle term of -11; ie -9 x -2:

`therefore 0= (q-2)(q-9)`

`therefore q=2` or `q=9`

Solving for p(pq=18) or (p+q=11) shows that p and q are interchangeable:

**Ans:**

**Therefore when q=2, p=9 and when q=9, p=2 **

`(p-q)^2=(p+q)^2-4pq`

`=11^2-4*18`

`=121-72`

`=49`

Hence, `p-q=+-7`

Taking p-q=7

and p+q=11

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Upon solving, p=9 and q=2

Again taking p-q=-7

and p+q=11

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Upon solving, p=2 and q=9

Therefore, the solutions for (p,q) are (9,2) or (2,9).