One linear one quadratic — substitution — Mathematics

Before this

This leads to

Simultaneous equations word problems
The equation of a circle · coming soon
Tangents and normals to a circle · coming soon

A straight line meets a curve

Solve these two equations together:

y = 2 x + 1 y = x^{2} + 3 x - 1

Try $x = 1$ . The first equation gives $y = 3$ . The second gives $y = 1 + 3 - 1 = 3$ . Both agree, so $(1, 3)$ is a solution.

Now try $x = - 2$ . The first equation gives $y = - 4 + 1 = - 3$ . The second gives $y = 4 - 6 - 1 = - 3$ . They agree again. So $(- 2, - 3)$ is also a solution.

This breaks something you have relied on for two lectures. Two linear equations have one solution, or none, or infinitely many. They never have exactly two. Yet this pair has exactly two, and both are correct. The reason is the second equation. It is not a straight line. It is a parabola, and a straight line can cross a parabola twice.

So the questions change. How many solutions can a pair like this have? How do we find all of them, without guessing? And how do we know we have not missed one? This lecture answers all three. The method is one you already know: substitution.

How many times can a line meet a parabola?

Before any algebra, look at the picture. Every solution of a pair of equations is a point that lies on both graphs. So the number of solutions is the number of points where the graphs meet.

Take the parabola $y = x^{2} + 3 x - 1$ and slide a straight line across it, always keeping the same gradient. Three things can happen.

Three parallel lines of gradient 2 against the parabola y = x² + 3x − 1. The solid line crosses it at (−2, −3) and (1, 3). The dashed line touches it once, at (−0.5, −2.25). The dotted line misses it.

If the line cuts through the curve, entering on one side and leaving on the other, it meets the curve twice. If it just touches the curve at one point, it meets it once. A line like this is called a tangent to the curve. If the line passes below the curve, it never meets it at all.

So a line and a parabola can have two solutions, one solution, or none. There is no fourth case. A parabola bends one way only, so a straight line cannot enter and leave it more than once. The algebra below will prove this, and it will also tell us which case we are in.

The method: make the two y's equal

Go back to the opening pair. Both equations say what $y$ is. At a solution, the same $y$ makes both true. So the two expressions for $y$ must be equal there:

2 x + 1 = x^{2} + 3 x - 1

This is substitution, exactly as in the first lecture of this topic. We replaced $y$ in the second equation by something equal to it, $2 x + 1$ . Now there is one equation with one unknown.

But look at what kind of equation it is. It contains $x^{2}$ , so it is a quadratic. A quadratic is solved by putting everything on one side, so that the other side is zero. Take $2 x + 1$ away from both sides:

00 = x^{2} + 3 x - 1 - 2 x - 1 = x^{2} + x - 2

Collect the terms carefully: $3 x - 2 x = x$ and $- 1 - 1 = - 2$ . Now factorise. We need two numbers that multiply to −2 and add to +1. They are +2 and −1:

(x + 2) (x - 1) = 0

So $x = - 2$ or $x = 1$ .

Here are the two solutions, appearing from the algebra. Look at what just happened. The quadratic has at most two roots, so the pair has at most two solutions. That is the proof that a line cannot meet this parabola three times.

The work is not finished. Each $x$ needs its own $y$ . Substitute each value into the linear equation, $y = 2 x + 1$ :

When $x = - 2$ : $y = 2 (- 2) + 1 = - 3$ .
When $x = 1$ : $y = 2 (1) + 1 = 3$ .

So the solutions are $(- 2, - 3)$ and $(1, 3)$ . These are the two points where the solid line crosses the parabola in the figure. The algebra and the picture tell the same story.

One linear and one quadratic equation

1. Make one letter the subject of the linear equation.

2. Substitute it into the quadratic equation, and rearrange to get a quadratic equal to zero.

3. Solve the quadratic, by factorising or by the formula. It gives two, one or no real values.

4. Substitute each value into the linear equation to find its partner.

5. Write the answers as pairs, and check each pair in both equations.

The discriminant decides how many

Step 3 produces a quadratic $a x^{2} + b x + c = 0$ . In the lecture on the discriminant, you saw that $b^{2} - 4 a c$ tells you how many roots a quadratic has, before you solve it. We call it $D$ . Some books write $Δ$ .

Here, every root of that quadratic is the $x$ -coordinate of a meeting point. So the discriminant now counts meeting points:

Discriminant	Roots of the quadratic	What the line does
$D > 0$	two different roots	crosses the curve at two points
$D = 0$	one repeated root	touches the curve once: it is a tangent
$D < 0$	no real roots	misses the curve

For the opening pair, $x^{2} + x - 2 = 0$ gives $D = 1^{2} - 4 (1) (- 2) = 9$ . This is positive, so there are two points, as we found.

Try it yourself on the graph below. The line keeps the gradient 2, but you can raise or lower it. The graph builder forms the quadratic for each position and works out its discriminant.

Graph builder — slide a line across a parabola

Move the line y = 2x + c up and down by changing c. Watch the discriminant. Find the exact value of c where the line touches the curve.

c = 1

Substituting y = 2x + c into y = x² + 3x − 1 gives x² + x − (1 + c) = 0. Its discriminant is D = 1 + 4(1 + c) = 5 + 4c.

c	Quadratic, = 0	D	Meeting points
1	x² + x − 2	9	two: (−2, −3) and (1, 3)
0	x² + x − 1	5	two: x ≈ 0.62 and x ≈ −1.62
−1.25	x² + x + 0.25	0	one: the tangent point (−0.5, −2.25)
−3	x² + x + 2	−7	none

The figure above shows the lines for c = 1, c = −1.25 and c = −3.

What did you see? Every time you lowered the line, the discriminant fell by 1. While it was positive, there were two meeting points, moving closer together. At $c = - 1.25$ , the discriminant was exactly zero. The two points became one, and the line touched the curve at $(- 0.5, - 2.25)$ . Below that, the discriminant was negative, and the line missed the curve.

That is it. That is the whole idea. A tangent is the moment when two meeting points merge into one, and a repeated root is the algebra's way of recording that moment.

Stop and think

Can a straight line meet the parabola $y = x^{2}$ at exactly one point without being a tangent to it?

Show the answer

Yes. The vertical line $x = 3$ meets $y = x^{2}$ only at $(3, 9)$ , but it cuts straight across the curve. It does not just touch it. Substitution shows why this case is different. Putting $x = 3$ into $y = x^{2}$ gives $y = 9$ at once. No quadratic appears, so there is no discriminant to look at. For every line that is not vertical, substitution gives a quadratic, and "exactly one meeting point" then means $D = 0$ , which means a tangent.

Worked examples

Worked Example 1 — When the quadratic will not factorise

Solve $y = x - 3$ and $y = x^{2} - 5 x + 3$ .

Both equations give $y$ , so set them equal: $x - 3 = x^{2} - 5 x + 3$ . Move everything to the right side: $0 = x^{2} - 6 x + 6$ .

Does this factorise? We need two numbers that multiply to 6 and add to −6. The factor pairs of 6 are 1 and 6, and 2 and 3. No pair adds to 6, so the quadratic does not factorise with integers. First check that roots exist: $D = (- 6)^{2} - 4 (1) (6) = 36 - 24 = 12$ . This is positive, so there are two meeting points. Use the formula:

x = \frac{6 \pm 12}{2} = \frac{6 \pm 2 3}{2} = 3 \pm 3

Here $12 = 4 \times 3 = 23$ . Now find each partner from the linear equation $y = x - 3$ :

When $x = 3 + 3$ : $y = 3$ .
When $x = 3 - 3$ : $y = - 3$ .

The solutions are $(3 + 3, 3)$ and $(3 - 3, - 3)$ . Leave them exact unless the question asks for decimals. As decimals, they are about $(4.73, 1.73)$ and $(1.27, - 1.73)$ .

Check the first pair in the quadratic: $(3 + 3)^{2} = 9 + 63 + 3 = 12 + 63$ . Then $12 + 63 - 15 - 53 + 3 = 3$ ✓.

Worked Example 2 — A line that touches

Solve $y = 2 x - 1$ and $y = x^{2}$ .

Set them equal: $x^{2} = 2 x - 1$ , so $x^{2} - 2 x + 1 = 0$ . The discriminant is $D = 4 - 4 = 0$ , so there is one repeated root. Factorise: $(x - 1)^{2} = 0$ , so $x = 1$ . Then $y = 2 (1) - 1 = 1$ .

There is one solution, $(1, 1)$ . The line $y = 2 x - 1$ is a tangent to $y = x^{2}$ at that point. Check: $1^{2} = 1$ ✓ and $2 (1) - 1 = 1$ ✓.

A single answer is not a sign that you lost one. When $D = 0$ , one answer is the complete answer, and the words "the line is a tangent" explain it.

Worked Example 3 — No meeting point at all

Solve $y = 3 - 2 x$ and $y = x^{2} - 4 x + 7$ .

Set them equal: $3 - 2 x = x^{2} - 4 x + 7$ . Bring everything to the right: $0 = x^{2} - 2 x + 4$ .

The discriminant is $D = (- 2)^{2} - 4 (1) (4) = 4 - 16 = - 12$ . It is negative, so the quadratic has no real roots. The line and the parabola never meet. The answer is: there are no real solutions.

This result often feels like an error, so check it another way. The quadratic is the curve's height minus the line's height. Complete the square: $x^{2} - 2 x + 4 = (x - 1)^{2} + 3$ . A square is never negative, so this difference is always at least 3. The parabola is always at least 3 units above the line. The closest they come is at $x = 1$ , where the curve is at 4 and the line is at 1.

Worked Example 4 — A line and a circle

Solve $x - y = - 2$ and $x^{2} + y^{2} = 10$ .

The second equation is a circle with centre $(0, 0)$ and radius $10$ . It is quadratic because it contains $x^{2}$ and $y^{2}$ . The method does not change.

Step 1: make $y$ the subject of the linear equation. Add $y$ and 2 to both sides: $y = x + 2$ .

Step 2: substitute into the circle. Use a bracket, because the whole of $x + 2$ is being squared:

x^{2} + (x + 2)^{2} = 10

Expand: $x^{2} + x^{2} + 4 x + 4 = 10$ , so $2 x^{2} + 4 x - 6 = 0$ . Divide every term by 2: $x^{2} + 2 x - 3 = 0$ .

Step 3: factorise. $(x + 3) (x - 1) = 0$ , so $x = - 3$ or $x = 1$ .

Step 4: use the linear equation, $y = x + 2$ . When $x = - 3$ , $y = - 1$ . When $x = 1$ , $y = 3$ .

The solutions are $(- 3, - 1)$ and $(1, 3)$ . Check both in the circle: $9 + 1 = 10$ ✓ and $1 + 9 = 10$ ✓. Check both in the line: $- 3 - (- 1) = - 2$ ✓ and $1 - 3 = - 2$ ✓.

Why the partner comes from the linear equation

Step 4 always uses the linear equation. The usual reason given is that it is simpler. That is true, but there is a deeper reason, and the circle shows it.

A non-vertical straight line has exactly one point above each $x$ . So each value of $x$ gives exactly one $y$ on the line. A circle is different. Above most values of $x$ it has two points, one on the top half and one on the bottom half. So if you put $x = 1$ into $x^{2} + y^{2} = 10$ , you get $y^{2} = 9$ , which gives $y = 3$ or $y = - 3$ . Only one of those points is on the line. The other is a false answer.

The linear equation gives the right partner automatically, because every solution must lie on the line. Test this on the working below.

Spot the error — a line and a circle

A student solves x + y = 7 and x² + y² = 25. Exactly one line contains the mistake. Tap it.

A student's working for x + y = 7 and x² + y² = 25. Line 5 is wrong.

Line 1. From the linear equation, y = 7 − x. Correct.
Line 2. Substitute: x² + (7 − x)² = 25. Correct.
Line 3. Expand: 2x² − 14x + 49 = 25, so x² − 7x + 12 = 0. Correct.
Line 4. Factorise: (x − 3)(x − 4) = 0, so x = 3 or x = 4. Correct.
Line 5. Put each x into x² + y² = 25: x = 3 gives y = ±4, and x = 4 gives y = ±3. So there are four solutions. Wrong: (3, −4) and (4, −3) are on the circle but not on the line.

The correct solutions are (3, 4) and (4, 3), found from y = 7 − x.

The four "solutions" in line 5 include two points that lie on the circle but not on the line. A pair must satisfy both equations. That is why the final check, in both equations, matters so much here.

Three classic mistakes

Trap 1 — Writing the answers as a list instead of pairs

" $x = - 2$ or $x = 1$ , $y = - 3$ or $y = 3$ " suggests four combinations. But $(- 2, 3)$ is not a solution: the line gives $y = 2 (- 2) + 1 = - 3$ when $x = - 2$ . Write each solution as a pair, $(- 2, - 3)$ and $(1, 3)$ , or label them clearly as "Solution 1" and "Solution 2".

Trap 2 — Stopping at the x values

The quadratic gives only the $x$ -coordinates. A solution of a pair of equations is a point, so it needs both coordinates. Every $x$ needs its partner $y$ from the linear equation.

Trap 3 — Treating "no real solutions" as a mistake

When the discriminant is negative, many students go back and change their working until it is positive. If your algebra is right, a negative discriminant is the answer: the graphs do not meet. Check the arithmetic once, then write "no real solutions" with the value of $D$ as the reason.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Solve $y = x + 4$ and $y = x^{2} - 2$ .

Hint 1
Both equations give $y$ . Set the right sides equal.

Hint 2
You should reach $x^{2} - x - 6 = 0$ . Find two numbers that multiply to −6 and add to −1.

Full answer
$x + 4 = x^{2} - 2$ gives $x^{2} - x - 6 = 0$ , so $(x - 3) (x + 2) = 0$ and $x = 3$ or $x = - 2$ . From $y = x + 4$ : the pairs are $(3, 7)$ and $(- 2, 2)$ . Check in the quadratic: $9 - 2 = 7$ ✓ and $4 - 2 = 2$ ✓.
Solve $x + 2 y = 7$ and $x y = 6$ .

Hint 1
Making $y$ the subject gives a fraction. Making $x$ the subject does not.

Hint 2
$x = 7 - 2 y$ . Substitute: $(7 - 2 y) y = 6$ , so $2 y^{2} - 7 y + 6 = 0$ .

Full answer
$2 y^{2} - 7 y + 6 = 0$ factorises as $(2 y - 3) (y - 2) = 0$ , so $y = \frac{3}{2}$ or $y = 2$ . From $x = 7 - 2 y$ : when $y = \frac{3}{2}$ , $x = 4$ ; when $y = 2$ , $x = 3$ . The pairs are $(4, \frac{3}{2})$ and $(3, 2)$ . Check: $4 \times \frac{3}{2} = 6$ ✓, $3 \times 2 = 6$ ✓, $4 + 3 = 7$ ✓, $3 + 4 = 7$ ✓. The method works for any letter. Choose the one that is easiest to isolate.
Solve $y = x + 1$ and $x^{2} + y^{2} = 13$ .

Hint 1
Substitute $x + 1$ for $y$ inside a bracket: $x^{2} + (x + 1)^{2} = 13$ .

Hint 2
You should reach $2 x^{2} + 2 x - 12 = 0$ . Divide by 2 first.

Full answer
$x^{2} + x^{2} + 2 x + 1 = 13$ gives $x^{2} + x - 6 = 0$ , so $(x + 3) (x - 2) = 0$ and $x = - 3$ or $x = 2$ . From $y = x + 1$ : the pairs are $(- 3, - 2)$ and $(2, 3)$ . Check in the circle: $9 + 4 = 13$ ✓ and $4 + 9 = 13$ ✓. Notice that $(2, - 3)$ is also on the circle, but it is not on the line, which is why the partner comes from the line.
The graph builder showed that the line $y = 2 x + c$ touches the curve $y = x^{2} + 3 x - 1$ for one value of $c$ . Find that value exactly, and the point where it touches. For which values of $c$ does the line cross the curve twice?

Hint 1
Substitute and collect terms, keeping $c$ as a letter. You should get $x^{2} + x - 1 - c = 0$ .

Hint 2
Touching means one repeated root, so $D = 0$ . Here $a = 1$ , $b = 1$ and the constant term is $- 1 - c$ .

Full answer
$D = 1^{2} - 4 (1) (- 1 - c) = 1 + 4 + 4 c = 5 + 4 c$ . Touching needs $5 + 4 c = 0$ , so $c = - \frac{5}{4} = - 1.25$ . The quadratic becomes $x^{2} + x + \frac{1}{4} = 0$ , which is $(x + \frac{1}{2})^{2} = 0$ , so $x = - \frac{1}{2}$ . Then $y = 2 (- \frac{1}{2}) - \frac{5}{4} = - \frac{9}{4}$ . The point is $(- \frac{1}{2}, - \frac{9}{4})$ . Check on the curve: $\frac{1}{4} - \frac{3}{2} - 1 = - \frac{9}{4}$ ✓. Two crossing points need $D > 0$ , so $5 + 4 c > 0$ , which means $c > - 1.25$ . This matches the graph builder: the discriminant fell by 1 for every 0.25 you lowered the line, and reached zero at $c = - 1.25$ .

The bigger idea

Count the solutions in this topic so far. Two straight lines meet at most once. A straight line and a parabola meet at most twice. The pattern comes from the equations. Substituting a line into a quadratic gives a quadratic, and a quadratic has at most two roots.

The pattern continues. Two parabolas of the form $y = a x^{2} + b x + c$ also meet at most twice, because subtracting one from the other leaves at most a quadratic. But two curves that are both quadratic in $x$ and $y$ , such as a circle and a sideways parabola, can meet at up to four points. A general rule says that a curve whose equation has degree $m$ and one of degree $n$ meet at no more than $m \times n$ points. It is called Bézout's theorem, after the French mathematician Étienne Bézout, who published a proof in 1779.

So an algebra question, "how many solutions?", is also a geometry question, "how many meeting points?". The discriminant answers both at once. You will use this again in Pre-University work, where you find the tangents to a circle by asking when a quadratic has $D = 0$ .

Key concepts in this lesson

Each solution of a pair of equations is a point where their graphs meet, so a line and a parabola can have two, one or no real solutions.
To solve one linear and one quadratic equation, make a letter the subject of the linear equation and substitute it into the quadratic.
The quadratic that results has a discriminant, and $D > 0$ , $D = 0$ and $D < 0$ mean two meeting points, a tangent, and no meeting point.
Each partner value comes from the linear equation, because a curve such as a circle can have two points above one value of $x$ .
Solutions are written as pairs, such as $(- 2, - 3)$ and $(1, 3)$ , and each pair is checked in both equations.