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.