Solve the simultaneous equations y - 3x + 2 = 0 y^2 - x - 6x^2 = 0 - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 2

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

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

Expanding the term: $9x^2 - 12x + 4 - x - 6x^2 = 0$ Combining like terms results in: $3x^2 - 13x + 4 = 0$

To solve the quadratic equation, we apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where:

• $a = 3$
• $b = -13$
• $c = 4$

Calculating the discriminant: $b^2 - 4ac = (-13)^2 - 4(3)(4) = 169 - 48 = 121$

Thus, we have: $x = \frac{13 \pm \sqrt{121}}{6} = \frac{13 \pm 11}{6}$ This gives us two potential solutions for x:

1. For the positive case: $x = \frac{24}{6} = 4$
2. For the negative case: $x = \frac{2}{6} = \frac{1}{3}$

Now we need to find the corresponding y-values for each x:

1. If $x = 4$: $y = 3(4) - 2 = 12 - 2 = 10$ So one solution pair is $(x, y) = (4, 10)$.

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

Thus, the solutions to the simultaneous equations are:

1. $(4, 10)$
2. $(\frac{1}{3}, -1)$