I don't approximate. To me that's like turning a correct answer into a wrong answer.
[tex]y = 3x+5[/tex]
[tex]y = 4x^2 + x[/tex]
[tex]3x + 5 = 4x^2 + x[/tex]
[tex]0 = 4x^2 - 2x - 5[/tex]
[tex]x = \frac 1 4 (1 \pm \sqrt{1^2 - (4)(-5)})[/tex]
[tex]x = \frac 1 4 (1 \pm \sqrt{21})[/tex]
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:
[tex]y_+ = 3x_++5 = \frac 3 4(1 + \sqrt{21}) + 5 = \frac 1 4(23 + 3\sqrt{21})[/tex]
[tex]y_- = 3x_- +5 = \frac 3 4(1 - \sqrt{21}) + 5 = \frac 1 4(23 - 3\sqrt{21})[/tex]
Answer:
[tex](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) [/tex]
You're on your own for the calculate work.
