Solve the simultaneous equations $$ y - 3x + 2 = 0 $$ $$ y^2 - x - 6x^2 = 0 $$ - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 2

Next, we substitute this expression for y into the second equation:

(3x - 2)^2 - x - 6x^2 = 0$$

Now, expand the squared term:

9x^2 - 12x + 4 - x - 6x^2 = 0$$ This simplifies to:

3x^2 - 13x + 4 = 0$$

We can use the quadratic formula to find x:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For our equation: - a = 3, b = -13, c = 4. Plugging in these values:

x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} = \frac{13 \pm \sqrt{169 - 48}}{6} = \frac{13 \pm \sqrt{121}}{6}$$ Calculating gives us:

x = \frac{13 \pm 11}{6}$$ So the possible values for x are: - $$x = 4$$ - $$x = \frac{1}{3}$$

Now we substitute back to find the corresponding y values:

1. If $x = 4$: $y = 3(4) - 2 = 10$ Thus, one solution is (4, 10).

2. If $x = \frac{1}{3}$: $y = 3 \left(\frac{1}{3}\right) - 2 = 1 - 2 = -1$ Thus, the other solution is (\left(\frac{1}{3}, -1\right)).

The final solutions to the simultaneous equations are:

• (4, 10)
• (\left(\frac{1}{3}, -1\right))