(a) Solve the equation, correct to two decimal places \(2x^{2} + 7x - 11 = 0\) (b) Using the substitution \(P = \frac{1}{x}; Q = \frac{1}{y}\), solve the si...

(a) Solve \(2x^2 + 7x - 11 = 0\) using the quadratic formula with \(a = 2, b = 7, c = -11\).

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 88}}{4} = \frac{-7 \pm \sqrt{137}}{4}.\]

Now \(\sqrt{137} = 11.7047\), so

\[x = \frac{-7 + 11.7047}{4} = \frac{4.7047}{4} = 1.18,\qquad x = \frac{-7 - 11.7047}{4} = \frac{-18.7047}{4} = -4.68.\]

Hence \(x = 1.18\) or \(x = -4.68\) (2 d.p.).

(b) Let \(P = \dfrac{1}{x}\) and \(Q = \dfrac{1}{y}\). The equations become:

\[2P + Q = 3 \quad(1), \qquad P - 5Q = 7 \quad(2).\]

From (1): \(Q = 3 - 2P\). Substitute into (2):

\[P - 5(3 - 2P) = 7 \Rightarrow P - 15 + 10P = 7 \Rightarrow 11P = 22 \Rightarrow P = 2.\]

Then \(Q = 3 - 2(2) = -1\).

Returning to \(x\) and \(y\): \(x = \dfrac{1}{P} = \dfrac{1}{2}\) and \(y = \dfrac{1}{Q} = -1\).

Therefore \(x = \dfrac{1}{2},\; y = -1\).