Simultaneous linear and quadratic equations

AS · P1 · 11 min

When one equation is linear and the other is quadratic, you solve them together by substitution: use the linear equation to replace one variable in the quadratic, which leaves an ordinary quadratic equation in one variable. Geometrically you are finding where a straight line meets a curve, which is why this skill runs through coordinate geometry, circles, tangents and area questions. It is usually worth 4 to 5 marks, and the last two of them are for pairing the coordinates correctly.

Why substitution works

Think of each equation as a graph. A solution of the pair is a point $(x, y)$ that lies on both graphs at once: an intersection point.

The linear equation says $y$ (or $x$ ) is a simple expression in the other variable. Replacing $y$ by that expression in the quadratic equation keeps only the points on the line, and asks which of them also lie on the curve. The result is a quadratic in one variable, so there are at most two solutions, matching the fact that a line meets a parabola or a circle at most twice.

Key result

Substituting the linear equation into the quadratic one gives a single quadratic equation. Its number of real roots is the number of intersection points:

two distinct roots: the line crosses the curve at two points;
one repeated root: the line touches the curve (it is a tangent);
no real roots: the line misses the curve.

Each root gives one value of the variable you kept; substitute it back into the linear equation to find the matching value of the other variable.

y = x^2 - 4x + 5 y = 2x - 3 (2, 1) (4, 5)

Solving one linear and one quadratic equation

Rearrange the linear equation to make one variable the subject. Choose the variable that avoids fractions if you can.
Substitute this expression into the quadratic equation. Put it in brackets.
Expand, collect terms, and rearrange to the form $ax^2 + bx + c = 0$ .
Solve the quadratic (factorise, or use the formula).
Substitute each root into the linear equation to find the other coordinate.
Write the answers as pairs, for example $x = 2,\ y = 1$ and $x = 4,\ y = 5$ , or as points $(2, 1)$ and $(4, 5)$ .

Substituting back into the linear equation is quicker and safer than using the quadratic one. The quadratic equation can give extra, false partners: from $x^2 + y^2 = 25$ with $x = 3$ , you get $y = \pm 4$ , but only one of these is on the line.

A line and a parabola

Solve the simultaneous equations $y = 2x - 3$ and $y = x^2 - 4x + 5$ .

Solution

The linear equation already gives $y$ . Substitute:

2x - 3 = x^2 - 4x + 5 \quad\Rightarrow\quad x^2 - 6x + 8 = 0 \quad\Rightarrow\quad (x - 2)(x - 4) = 0

So $x = 2$ or $x = 4$ . Using $y = 2x - 3$ : when $x = 2$ , $y = 1$ ; when $x = 4$ , $y = 5$ .

Solutions: $x = 2,\ y = 1$ and $x = 4,\ y = 5$ .

A line and a circle

Solve the simultaneous equations $x + y + 1 = 0$ and $x^2 + y^2 = 25$ .

Solution

From the linear equation, $y = -x - 1$ . Substitute:

\begin{aligned} x^2 + (-x - 1)^2 &= 25 \\ x^2 + x^2 + 2x + 1 &= 25 \\ 2x^2 + 2x - 24 &= 0 \\ x^2 + x - 12 &= 0 \\ (x + 4)(x - 3) &= 0 \end{aligned}

So $x = -4$ or $x = 3$ . Then $y = -x - 1$ gives $y = 3$ and $y = -4$ respectively.

Solutions: $(-4, 3)$ and $(3, -4)$ .

Note $(-x - 1)^2 = (x + 1)^2 = x^2 + 2x + 1$ . Squaring a bracket with a minus sign inside is a frequent source of errors; write the bracket out in full.

Choosing the variable to eliminate

Solve the simultaneous equations $x + 2y = 7$ and $xy = 6$ .

Solution

Making $x$ the subject avoids fractions: $x = 7 - 2y$ . Substitute into $xy = 6$ :

(7 - 2y)y = 6 \quad\Rightarrow\quad 7y - 2y^2 = 6 \quad\Rightarrow\quad 2y^2 - 7y + 6 = 0 \quad\Rightarrow\quad (2y - 3)(y - 2) = 0

So $y = 2$ or $y = \tfrac{3}{2}$ . Then $x = 7 - 2y$ gives $x = 3$ and $x = 4$ .

Solutions: $x = 3,\ y = 2$ and $x = 4,\ y = \tfrac{3}{2}$ .

An equation with an xy term

Solve the simultaneous equations $2x + 3y = 7$ and $3x^2 = 4 + 4xy$ .

Solution

From the linear equation, $y = \dfrac{7 - 2x}{3}$ . (Making $x$ the subject would also work, but this keeps the $x^2$ term simple.) Substitute:

3x^2 = 4 + 4x \cdot \frac{7 - 2x}{3}

Multiply by $3$ to clear the fraction:

\begin{aligned} 9x^2 &= 12 + 28x - 8x^2 \\ 17x^2 - 28x - 12 &= 0 \end{aligned}

$ac = -204$ , and $-34 + 6 = -28$ , so $17x^2 - 34x + 6x - 12 = 17x(x - 2) + 6(x - 2) = (17x + 6)(x - 2) = 0$ .

So $x = 2$ or $x = -\dfrac{6}{17}$ .

$x = 2$ : $y = \dfrac{7 - 4}{3} = 1$ .
$x = -\dfrac{6}{17}$ : $y = \dfrac{7 + \frac{12}{17}}{3} = \dfrac{131}{51}$ .

Solutions: $(2, 1)$ and $\left(-\dfrac{6}{17}, \dfrac{131}{51}\right)$ .

If the factors are hard to spot, the formula gives $x = \dfrac{28 \pm \sqrt{784 + 816}}{34} = \dfrac{28 \pm 40}{34}$ , the same two values.

Midpoint and length of a chord

The line $x + y = 5$ meets the curve $x^2 + 2y^2 = 22$ at the points $A$ and $B$ . Find the coordinates of the midpoint of $AB$ and the exact length of $AB$ .

Solution

From the line, $x = 5 - y$ . Substitute:

\begin{aligned} (5 - y)^2 + 2y^2 &= 22 \\ 25 - 10y + y^2 + 2y^2 &= 22 \\ 3y^2 - 10y + 3 &= 0 \\ (3y - 1)(y - 3) &= 0 \end{aligned}

So $y = 3$ or $y = \tfrac{1}{3}$ , and $x = 5 - y$ gives $x = 2$ or $x = \tfrac{14}{3}$ . The points are $A(2, 3)$ and $B\left(\tfrac{14}{3}, \tfrac{1}{3}\right)$ .

Midpoint:

\left(\frac{2 + \frac{14}{3}}{2},\ \frac{3 + \frac{1}{3}}{2}\right) = \left(\frac{10}{3},\ \frac{5}{3}\right)

Length:

AB = \sqrt{\left(\tfrac{8}{3}\right)^2 + \left(-\tfrac{8}{3}\right)^2} = \sqrt{\tfrac{128}{9}} = \frac{8\sqrt{2}}{3}

x^2 + 2y^2 = 22 x + y = 5 (2, 3) (14/3, 1/3)

Showing that a line is a tangent

Show that the line $y = 3x - 2$ is a tangent to the curve $y = x^2 + x - 1$ , and find the point of contact.

Solution

Substitute:

3x - 2 = x^2 + x - 1 \quad\Rightarrow\quad x^2 - 2x + 1 = 0 \quad\Rightarrow\quad (x - 1)^2 = 0

The quadratic has a repeated root, so the line meets the curve at exactly one point: it is a tangent. The point of contact is $x = 1$ , $y = 3(1) - 2 = 1$ , i.e. $(1, 1)$ .

You could also show the discriminant is zero: $(-2)^2 - 4(1)(1) = 0$ . Either way, state the conclusion: "repeated root, so the line is a tangent".

Watch out

Substituting into the wrong equation. Rearrange the linear equation and substitute into the quadratic. The other way round produces a square root or a messier equation.

Unbracketed substitution. Writing $x^2 + -x - 1^2$ instead of $x^2 + (-x - 1)^2$ loses the cross term $2x$ . Always use brackets.

Mismatched pairs. Listing " $x = -4, 3$ and $y = 3, -4$ " without saying which goes with which can lose the final mark. Write the solutions as pairs.

Finding $y$ from the quadratic equation. This can produce extra values of $y$ that are not on the line. Use the linear equation.

Stopping at $x$ . "Solve the simultaneous equations" requires both variables.

Exam tip

Typical mark scheme: M1 for substituting to get an equation in one variable, A1 for the correct three-term quadratic, M1 for solving it, A1 for both $x$ -values, A1 for both $y$ -values (or both correct pairs).
The three-term quadratic is often "given" as a check in part (a) ("show that the $x$ -coordinates satisfy ..."). Show every line of the rearrangement; you cannot skip to the given answer.
If exact answers are required, leave surds in. Otherwise give coordinates to 3 significant figures.
Questions often continue: find the midpoint or length of the chord, or the equation of the perpendicular bisector of $AB$ . Keep the coordinates exact (fractions, surds) so later parts are accurate.

Summary

Rearrange the linear equation and substitute into the quadratic.
Choose the variable to eliminate so as to avoid fractions where possible.
Bracket the substituted expression, expand carefully, and rearrange to $ax^2 + bx + c = 0$ .
Find the other coordinate from the linear equation.
Present solutions as pairs or points.
Two distinct roots: two intersection points; repeated root: tangent; no real roots: no intersection.

Practice questions

Question

Solve the simultaneous equations $y = 3x + 1$ and $y = 2x^2 - x - 5$ .
Solve the simultaneous equations $x - y = 2$ and $x^2 + y^2 = 20$ .
Solve the simultaneous equations $2x + y = 5$ and $x^2 + xy = 6$ .
Solve the simultaneous equations $x + 3y = 5$ and $x^2 - 3xy + 2y^2 = 0$ .
Show that the line $y = x - 5$ does not meet the curve $y = x^2 - 3x$ .
The line $y = 2x + 1$ meets the curve $x^2 + y^2 = 13$ at $A$ and $B$ . Find the coordinates of $A$ and $B$ and the exact length of $AB$ .
The line $y = x + 2$ meets the curve $y = x^2 - 4x + 6$ at $P$ and $Q$ . Find the coordinates of the midpoint of $PQ$ and the equation of the perpendicular bisector of $PQ$ .
Solve the simultaneous equations $y = 2x - 3$ and $x^2 + y^2 - 6x + 4y = 12$ .
Solve the simultaneous equations $x + y = 6$ and $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}$ .

Answers

$3x + 1 = 2x^2 - x - 5$ gives $2x^2 - 4x - 6 = 0$ , i.e. $x^2 - 2x - 3 = 0$ , $(x - 3)(x + 1) = 0$ . Points $(3, 10)$ and $(-1, -2)$ .
$x = y + 2$ : $(y + 2)^2 + y^2 = 20$ , so $2y^2 + 4y - 16 = 0$ , $y^2 + 2y - 8 = 0$ , $(y + 4)(y - 2) = 0$ . $y = 2$ , $x = 4$ ; $y = -4$ , $x = -2$ . Solutions $(4, 2)$ and $(-2, -4)$ .
$y = 5 - 2x$ : $x^2 + x(5 - 2x) = 6$ , so $-x^2 + 5x - 6 = 0$ , $x^2 - 5x + 6 = 0$ , $x = 2$ or $3$ . Solutions $(2, 1)$ and $(3, -1)$ .
$x = 5 - 3y$ : $(5 - 3y)^2 - 3y(5 - 3y) + 2y^2 = 0$ , so $25 - 30y + 9y^2 - 15y + 9y^2 + 2y^2 = 0$ , i.e. $20y^2 - 45y + 25 = 0$ , $4y^2 - 9y + 5 = 0$ , $(4y - 5)(y - 1) = 0$ . $y = 1$ , $x = 2$ ; $y = \tfrac{5}{4}$ , $x = \tfrac{5}{4}$ . Solutions $(2, 1)$ and $\left(\tfrac{5}{4}, \tfrac{5}{4}\right)$ .
$x - 5 = x^2 - 3x$ gives $x^2 - 4x + 5 = 0$ . Discriminant $16 - 20 = -4 < 0$ , so no real roots: the line does not meet the curve.
$x^2 + (2x + 1)^2 = 13$ gives $5x^2 + 4x - 12 = 0$ , $(5x - 6)(x + 2) = 0$ . $A(-2, -3)$ , $B\left(\tfrac{6}{5}, \tfrac{17}{5}\right)$ . $AB = \sqrt{\left(\tfrac{16}{5}\right)^2 + \left(\tfrac{32}{5}\right)^2} = \tfrac{16}{5}\sqrt{1 + 4} = \tfrac{16\sqrt{5}}{5}$ .
$x + 2 = x^2 - 4x + 6$ gives $x^2 - 5x + 4 = 0$ , so $x = 1$ or $4$ : $P(1, 3)$ , $Q(4, 6)$ . Midpoint $\left(\tfrac{5}{2}, \tfrac{9}{2}\right)$ . Gradient of $PQ$ is $1$ , so the perpendicular gradient is $-1$ : $y - \tfrac{9}{2} = -\left(x - \tfrac{5}{2}\right)$ , i.e. $y = -x + 7$ .
$x^2 + (2x - 3)^2 - 6x + 4(2x - 3) = 12$ gives $5x^2 - 10x - 15 = 0$ , $x^2 - 2x - 3 = 0$ , $x = 3$ or $-1$ . Solutions $(3, 3)$ and $(-1, -5)$ .
$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x + y}{xy} = \dfrac{6}{xy}$ , so $\dfrac{6}{xy} = \dfrac{3}{4}$ and $xy = 8$ . With $y = 6 - x$ : $x(6 - x) = 8$ , so $x^2 - 6x + 8 = 0$ , $x = 2$ or $4$ . Solutions $(2, 4)$ and $(4, 2)$ .

Why substitution works

Practice questions

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