A Linear and a Quadratic Together
Sometimes one of two simultaneous equations is a straight line and the other is a curve. For
example, take a line and a parabola:
y = x + 1 \qquad\text{and}\qquad y = x^2.
A solution is a pair (x, y) that fits both rules at once —
and just as with two lines, that is exactly where their graphs
cross. Because one of them bends, a line and a parabola can meet in
two places, not just one.
Substitute the line into the curve
The
substitution method
works just as before. Both equations already give y on their own,
so set the two expressions for y equal:
x^2 = x + 1.
Notice what happened: the y is gone, and we are left with a single
equation in x alone. Move everything to one side and it is an
ordinary quadratic:
x^2 - x - 1 = 0.
That is the heart of the method: substituting the linear equation into the quadratic
leaves a quadratic in one variable, which we already know how to solve by
factorising.
A case that factorises cleanly
The general power of this method is clear, but let us pick a pair that factorises neatly so we
can read off whole-number answers. Take the line y = x + 6 and the
parabola y = x^2. Setting them equal and tidying up:
x^2 = x + 6 \;\Longrightarrow\; x^2 - x - 6 = 0 \;\Longrightarrow\; (x - 3)(x + 2) = 0.
So x = 3 or x = -2. Each
x gives its partner y when we put it back
into the linear equation (the easier one):
x = 3 \Rightarrow y = 9, \qquad x = -2 \Rightarrow y = 4.
The two solutions are (3, 9) and
(-2, 4) — the two points where this line cuts the curve.
See the two crossings
Here are y = x + 1 and y = x^2 on one set
of axes. The line cuts the
parabola
at two points — those crossings are precisely the two
(x, y) pairs that solve the system (the solutions of
x^2 - x - 1 = 0).
A line can also just graze a parabola at one point (a tangent), or
miss it entirely — matching a quadratic with one repeated root, or no real
roots at all.