0606 P23 - Nov 2021 - Q2 - 5 marks
8064
Solve the simultaneous equations. Give your answers in the form \(a+b\sqrt3\), where \(a\) and \(b\) are rational.
\(x+y=3,\qquad 2x-\sqrt3\,y=5.\)
Solution
Answer: \(x=1+\sqrt3,\ y=2-\sqrt3\).
The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.
From
\(x+y=3,\)
we have
\(y=3-x.\)
Substitute this into the second equation:
\(2x-\sqrt3(3-x)=5.\)
Expand:
\(2x-3\sqrt3+\sqrt3x=5.\)
So
\(x(2+\sqrt3)=5+3\sqrt3.\)
Hence
\(x=\frac{5+3\sqrt3}{2+\sqrt3}.\)
Rationalise the denominator:
\(x=\frac{(5+3\sqrt3)(2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)}.\)
The denominator is \(1\), and the numerator is
\(10-5\sqrt3+6\sqrt3-9=1+\sqrt3.\)
Therefore
\(x=1+\sqrt3.\)
Then
\(y=3-(1+\sqrt3)=2-\sqrt3.\)
