Question

Solve the simultaneous equation:

Answer 1

I don't approximate.  To me that's like turning a correct answer into a wrong answer.

$y = 3x+5$

$y = 4x^2 + x$

$3x + 5 = 4x^2 + x$

$0 = 4x^2 - 2x - 5$

$x = \frac 1 4 (1 \pm \sqrt{1^2 - (4)(-5)})$

$x = \frac 1 4 (1 \pm \sqrt{21})$

So there are two values for x.  Each has an associated y.  Since our quadratics have integer coefficients, the pair of ys, like the pair of xs, will be quadratic conjugates:

$y_+ = 3x_++5 = \frac 3 4(1 + \sqrt{21}) + 5 = \frac 1 4(23 + 3\sqrt{21})$

$y_- = 3x_- +5 = \frac 3 4(1 - \sqrt{21}) + 5 = \frac 1 4(23 - 3\sqrt{21})$

Answer:

$(x,y) = \left( \frac 1 4 (1 + \sqrt{21}), \frac 1 4(23 + 3\sqrt{21})\right) \textrm{ or } \left( \frac 1 4 (1 - \sqrt{21}), \frac 1 4(23 - 3\sqrt{21})\right)$

You're on your own for the calculate work.