Throw a ball at a wall. The ball's path through the air is a curve — a parabola, arcing up and bending back down. The wall (or a sloping ramp, or a laser beam aimed across a satellite dish) is a straight line. The question "where does the ball hit the ramp?" is really the question "where does a line meet a curve?" — and that is exactly what this page teaches you to answer.

Until now, our simultaneous equations have been two straight lines, which cross at most once. The moment one equation bends, something new happens: a line and a parabola can meet at two points, just graze at one, or miss each other completely. Every one of those situations has an algebraic twin, and this page keeps both views in play the whole way through:

• Algebraically, we substitute one equation into the other and solve the quadratic that is left.
• Geometrically, the solutions are the crossing points of the two graphs — two, one, or none.

Neither view is the "real" one. The algebra tells you the exact coordinates; the picture tells you how many answers to expect and whether your answers are sensible. A confident mathematician flicks between the two constantly.

The setup

Take a line and a parabola:

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. Try a point and see: does (2, 3) work? It fits the line (3 = 2 + 1 ✓) but not the parabola (3 \ne 2^2 ✗) — so it sits on the line but off the curve, and it is not a solution. We need points that sit on both graphs at once.

Because one of the graphs bends, there can be two such points, not just one: the line can enter the parabola's "cup", cross it on the way in, and cross it again on the way out.

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:

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:

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 (or, when it refuses to factorise, with the quadratic formula). Everything you already know about quadratics — how many roots they have, when they have none — instantly becomes knowledge about where lines meet curves.

A full worked example, start to finish

Solve the pair

Step 1 — set the two y's equal. Both equations tell us what y is, so the two right-hand sides must match at any crossing point:

Step 2 — rearrange so one side is zero. Subtract x and 1 from both sides:

Step 3 — solve the quadratic. Two numbers that multiply to -2 and add to -1 are -2 and +1, so it factorises:

Step 4 — find each partner y. We are not done: x = 2 and x = -1 are only half of each answer. Put each one back into the linear equation y = x + 1:

Step 5 — state the solutions as pairs and check. The solutions are (2, 3) and (-1, 0) — not the loose numbers "2, -1, 3, 0". Checking in the other equation, the curve: 2^2 - 1 = 3 ✓ and (-1)^2 - 1 = 0 ✓. Both points sit on both graphs, so both are genuine crossings.

Another that factorises cleanly

One more for the pattern. Take the line y = x + 6 and the parabola y = x^2. Setting them equal and tidying up:

So x = 3 or x = -2. Each x gives its partner y when we put it back into the linear equation (the easier one):

The two solutions are (3, 9) and (-2, 4) — the two points where this line cuts the curve. Notice the rhythm, because it is always the same: set equal → rearrange to zero → solve the quadratic → back-substitute each x → answer in pairs.

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).

Look at where the crossings sit: one just left of x = 0, one a little past x = 1.6. This quadratic does not factorise with whole numbers — its exact roots are x = \frac{1 \pm \sqrt{5}}{2} — but the picture already told us there were exactly two of them, and roughly where. That is the geometric view earning its keep: even before you solve anything, the graph tells you how many answers to hunt for.

When the line only touches: a tangent

Now solve y = 2x - 1 with y = x^2. Same routine — set equal, rearrange, factorise:

The quadratic has a repeated root: x = 1, twice. There is only one crossing point, (1, 1) — and the picture shows why the root "repeats". The line does not cut through the parabola at all; it comes in, just touches it at a single point, and leaves on the same side. The two crossing points have merged into one. A line like this is called a tangent to the curve.

A repeated root in the algebra always means a tangent in the picture. This is the knife-edge case: nudge the line up a whisker and it cuts twice; nudge it down a whisker and it misses entirely.

When they never meet at all

Finally, try y = x - 2 with y = x^2:

This one refuses to factorise, and the quadratic formula tells us why. The discriminant — the part under the square root, b^2 - 4ac — comes out negative:

A negative number has no real square root, so the quadratic has no real solutions — and the graphs agree completely: the line passes below the parabola and never touches it. "They never meet" is not a failure of the algebra; it is the algebra working perfectly, reporting a true fact about the picture.

Try it: slide the line yourself

The parabola below is fixed at y = x^2; the line is y = x + c, and the slider moves it up and down. Substituting gives x^2 - x - c = 0, whose discriminant is \Delta = 1 + 4c. Watch all three situations happen as you slide:

Push c high and the crossings drift far apart (\Delta large). Bring it down to exactly c = -\tfrac14 — the slider snaps there — and the two crossings merge into a single touch (\Delta = 1 + 4(-\tfrac14) = 0: the tangent). Any lower and the line drops clear of the cup: no crossings, matching \Delta < 0. The slider is really a discriminant dial.

See it explained