Quadratic Simultaneous Equations

What you'll learn

How to solve two equations when one contains squared terms.
Why quadratic simultaneous equations often have two solutions.
How to substitute a linear equation into a quadratic equation.
How to handle non-factorising answers and round to 3 significant figures.

The big picture: line meets curve

Definition

Simultaneous equations

Simultaneous equations are two or more equations that must be true at the same time. A solution is usually an ordered pair, written like $(x, y)$ , that works in every equation.

In this topic, one equation is usually linear, meaning its graph is a straight line, and the other is quadratic, meaning it contains a squared term such as $x^2$ or $y^2$ .

A common example is:

x^2 + y^2 = 10

This is a circle centred at the origin. If you solve it together with a straight line, you are finding the points where the line and circle intersect.

Graph of a circle and straight line intersecting at two points

Key Idea

What the answers mean

Each solution is a coordinate pair. If there are two intersections, you should expect two coordinate pairs.

Example

Solving a line and circle by substitution

Solve:

x^2 + y^2 = 10

y = x + 2

The second equation already tells you what $y$ is, so substitute $y = x + 2$ into the quadratic equation:
$x^2 + (x + 2)^2 = 10$
Expand the bracket carefully:
$x^2 + x^2 + 4x + 4 = 10$
Collect like terms and rearrange to make the equation equal to zero:
$2x^2 + 4x - 6 = 0$
Divide every term by 2 to simplify:
$x^2 + 2x - 3 = 0$
Factorise the quadratic:
$(x + 3) (x - 1) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = 1$
Use $y = x + 2$ to find the matching $y$ -values:
$x=1,y=3\begin{aligned} \text{If } x &= -3, & y &= -1 \\ \text{If } x &= 1, & y &= 3 \end{aligned}$
Write the two coordinate pairs:
$\quad \text{and} \quad (1, 3)$

Common Mistake

Do not split the squares

From $x^2 + y^2 = 10$ , you cannot say $\sqrt{10}$ . Square roots do not work separately across addition.

The main method: substitute, solve, pair up

Definition

Substitution

Substitution means replacing one variable with an equal expression involving the other variable. This turns two equations into one equation with one unknown.

Most Grade 8/9 quadratic simultaneous questions follow this pattern:

Rearrange the linear equation if needed.
Substitute into the quadratic equation.
Solve the quadratic equation.
Substitute back to find the matching values.
Give answers as coordinate pairs.

Example

When the line is written as a difference

Solve:

x^2 + y^2 = 29

x - y = 3

Rearrange the linear equation to make $y$ the subject:
$\Rightarrow y = x - 3$
Substitute $y = x - 3$ into $x^2 + y^2 = 29$ :
$x^2 + (x - 3)^2 = 29$
Expand the bracket:
$x^2 + x^2 - 6x + 9 = 29$
Collect terms and rearrange:
$2x^2 - 6x - 20 = 0$
Divide by 2:
$x^2 - 3x - 10 = 0$
Factorise:
$(x - 5) (x + 2) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = -2$
Substitute into $y = x - 3$ :
$x=−2,y=−5\begin{aligned} \text{If } x &= 5, & y &= 2 \\ \text{If } x &= -2, & y &= -5 \end{aligned}$
Write the final answers:
$\quad \text{and} \quad (-2, -5)$

Tip

Pair the values correctly

Do not list all the $x$ -values and all the $y$ -values separately. Each $y$ -value comes from substituting a particular $x$ -value, so keep them as matched coordinate pairs.

Rearranging a line with coefficients

Sometimes the linear equation looks like $2 x + y = 5$ or $3 x = 4 - y$ . That is still just a straight line. Rearrange it before substituting.

Example

Line not already in the form $y = \dots$

Solve:

x^2 + y^2 = 25

4 x + y = 5

Rearrange the linear equation to make $y$ the subject:
$y = 5 - 4 x$
Substitute into the quadratic equation:
$x^2 + (5 - 4x)^2 = 25$
Expand the bracket:
$x^2 + 25 - 40x + 16x^2 = 25$
Simplify:
$17x^2 - 40x = 0$
Factorise by taking out the common factor $x$ :
$x (17 x - 40) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = \frac{40}{17}$
Substitute each value into $y = 5 - 4 x$ :
$x=4017,y=−7517\begin{aligned} \text{If } x &= 0, & y &= 5 \\ \text{If } x &= \frac{40}{17}, & y &= -\frac{75}{17} \end{aligned}$
Write the final answers:
$\quad \text{and} \quad \left(\frac{40}{17}, -\frac{75}{17}\right)$

When the quadratic does not factorise nicely

Definition

Quadratic formula

The quadratic formula solves equations of the form $ax^2 + bx + c = 0$ :

\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

If the question asks for answers to 3 significant figures, it is a strong hint that the quadratic may not factorise neatly.

Example

Answers to 3 significant figures

Solve, giving your answers to 3 significant figures:

x^2 + y^2 = 18

2 x + y = 4

Rearrange the linear equation:
$y = 4 - 2 x$
Substitute into the quadratic equation:
$x^2 + (4 - 2x)^2 = 18$
Expand and simplify:
$x2+16−16x+4x2=185x2−16x−2=0\begin{aligned} x^2 + 16 - 16x + 4x^2 &= 18 \\ 5x^2 - 16x - 2 &= 0 \end{aligned}$
Use the quadratic formula with $a = 5$ , $b = - 16$ , and $c = - 2$ :
$\frac{16 \pm \sqrt{(-16)^2 - 4(5)(-2)}}{10}$
Simplify inside the square root:
$\frac{16 \pm \sqrt{296}}{10}$
Find the two $x$ -values:
$\approx 3.320 \quad \text{or} \quad x \approx -0.120$
Substitute into $y = 4 - 2 x$ :
$x≈−0.120,y≈4.241\begin{aligned} \text{If } x &\approx 3.320, & y &\approx -2.641 \\ \text{If } x &\approx -0.120, & y &\approx 4.241 \end{aligned}$
Round both coordinate pairs to 3 significant figures:
$\quad \text{and} \quad (-0.120, 4.24)$

Tip

Rounding decimals less than 1

For a number like -0.120, the zero after the 2 is significant. Writing -0.12 gives only 2 significant figures, so keep -0.120 when 3 significant figures are required.

Quadratics that are not circles

Not every quadratic simultaneous equation looks like $x^2 + y^2 = k$ . You may see equations such as $x^2 - 3y^2 = 13$ or $2x^2 - y^2 = 14$ .

The method is the same: use the linear equation to replace one variable, then solve the resulting quadratic.

Definition

Solve algebraically

To solve algebraically means to use algebraic methods such as rearranging, substituting, expanding and factorising. You should not rely on a graph for your final answer.

Example

Solving a quadratic with different squared terms

Solve algebraically:

2x^2 - y^2 = 1

3 x + 2 y = 1

Rearrange the linear equation to make $y$ the subject:
$2 y = 1 - 3 x$
Divide by 2:
$\frac{1 - 3x}{2}$
Substitute into $2x^2 - y^2 = 1$ :
$2x2−(1−3x2)2=12x^2 - \left(\frac{1 - 3x}{2}\right)^2 = 1$
Multiply every term by 4 to remove the fraction:
$8x^2 - (1 - 3x)^2 = 4$
Expand the bracket carefully:
$8x^2 - (1 - 6x + 9x^2) = 4$
Remove the bracket and simplify:
$x^2 + 6x - 1 = 4$
Rearrange into standard quadratic form:
$x^2 - 6x + 5 = 0$
Factorise:
$(x - 1) (x - 5) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = 5$
Substitute into $\frac{1 - 3x}{2}$ :
$x=5,y=−7\begin{aligned} \text{If } x &= 1, & y &= -1 \\ \text{If } x &= 5, & y &= -7 \end{aligned}$
Write the two solutions:
$\quad \text{and} \quad (5, -7)$

Common Mistake

Forgetting the minus before a bracket

In expressions like $2x^2 - (1 - 3x)^2$ , the minus sign applies to the whole expanded bracket. Write the brackets out fully before simplifying.

Common Mistake

Sometimes there are not two real solutions

A line might touch the curve once or miss it completely. Algebraically, this shows up as a repeated root or a negative value under the square root in the quadratic formula.

Exam technique

In the exam

Rearrange the linear equation first, then substitute into the quadratic equation.
Keep your working neat when expanding brackets, especially when there is a minus sign before a squared bracket.
Always give answers as coordinate pairs and check each pair in both original equations if you have time.

Self review

Check yourself

Can you explain why a quadratic simultaneous equation can have two coordinate-pair solutions?
When substituting $y = 3 - 2 x$ into $x^2 + y^2 = 20$ , where do the brackets need to go?
If your quadratic gives two $x$ -values, how do you find the matching $y$ -values?

Quadratic Simultaneous Equations

What you'll learn

How to solve two equations when one contains squared terms.
Why quadratic simultaneous equations often have two solutions.
How to substitute a linear equation into a quadratic equation.
How to handle non-factorising answers and round to 3 significant figures.

The big picture: line meets curve

Definition

Simultaneous equations

Simultaneous equations are two or more equations that must be true at the same time. A solution is usually an ordered pair, written like $(x, y)$ , that works in every equation.

In this topic, one equation is usually linear, meaning its graph is a straight line, and the other is quadratic, meaning it contains a squared term such as $x^2$ or $y^2$ .

A common example is:

x^2 + y^2 = 10

This is a circle centred at the origin. If you solve it together with a straight line, you are finding the points where the line and circle intersect.

Graph of a circle and straight line intersecting at two points

Key Idea

What the answers mean

Each solution is a coordinate pair. If there are two intersections, you should expect two coordinate pairs.

Example

Solving a line and circle by substitution

Solve:

x^2 + y^2 = 10

y = x + 2

The second equation already tells you what $y$ is, so substitute $y = x + 2$ into the quadratic equation:
$x^2 + (x + 2)^2 = 10$
Expand the bracket carefully:
$x^2 + x^2 + 4x + 4 = 10$
Collect like terms and rearrange to make the equation equal to zero:
$2x^2 + 4x - 6 = 0$
Divide every term by 2 to simplify:
$x^2 + 2x - 3 = 0$
Factorise the quadratic:
$(x + 3) (x - 1) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = 1$
Use $y = x + 2$ to find the matching $y$ -values:
$x=1,y=3\begin{aligned} \text{If } x &= -3, & y &= -1 \\ \text{If } x &= 1, & y &= 3 \end{aligned}$
Write the two coordinate pairs:
$\quad \text{and} \quad (1, 3)$

Common Mistake

Do not split the squares

From $x^2 + y^2 = 10$ , you cannot say $\sqrt{10}$ . Square roots do not work separately across addition.

The main method: substitute, solve, pair up

Definition

Substitution

Substitution means replacing one variable with an equal expression involving the other variable. This turns two equations into one equation with one unknown.

Most Grade 8/9 quadratic simultaneous questions follow this pattern:

Rearrange the linear equation if needed.
Substitute into the quadratic equation.
Solve the quadratic equation.
Substitute back to find the matching values.
Give answers as coordinate pairs.

Example

When the line is written as a difference

Solve:

x^2 + y^2 = 29

x - y = 3

Rearrange the linear equation to make $y$ the subject:
$\Rightarrow y = x - 3$
Substitute $y = x - 3$ into $x^2 + y^2 = 29$ :
$x^2 + (x - 3)^2 = 29$
Expand the bracket:
$x^2 + x^2 - 6x + 9 = 29$
Collect terms and rearrange:
$2x^2 - 6x - 20 = 0$
Divide by 2:
$x^2 - 3x - 10 = 0$
Factorise:
$(x - 5) (x + 2) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = -2$
Substitute into $y = x - 3$ :
$x=−2,y=−5\begin{aligned} \text{If } x &= 5, & y &= 2 \\ \text{If } x &= -2, & y &= -5 \end{aligned}$
Write the final answers:
$\quad \text{and} \quad (-2, -5)$

Tip

Pair the values correctly

Do not list all the $x$ -values and all the $y$ -values separately. Each $y$ -value comes from substituting a particular $x$ -value, so keep them as matched coordinate pairs.

Rearranging a line with coefficients

Sometimes the linear equation looks like $2 x + y = 5$ or $3 x = 4 - y$ . That is still just a straight line. Rearrange it before substituting.

Example

Line not already in the form $y = \dots$

Solve:

x^2 + y^2 = 25

4 x + y = 5

Rearrange the linear equation to make $y$ the subject:
$y = 5 - 4 x$
Substitute into the quadratic equation:
$x^2 + (5 - 4x)^2 = 25$
Expand the bracket:
$x^2 + 25 - 40x + 16x^2 = 25$
Simplify:
$17x^2 - 40x = 0$
Factorise by taking out the common factor $x$ :
$x (17 x - 40) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = \frac{40}{17}$
Substitute each value into $y = 5 - 4 x$ :
$x=4017,y=−7517\begin{aligned} \text{If } x &= 0, & y &= 5 \\ \text{If } x &= \frac{40}{17}, & y &= -\frac{75}{17} \end{aligned}$
Write the final answers:
$\quad \text{and} \quad \left(\frac{40}{17}, -\frac{75}{17}\right)$

When the quadratic does not factorise nicely

Definition

Quadratic formula

The quadratic formula solves equations of the form $ax^2 + bx + c = 0$ :

\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

If the question asks for answers to 3 significant figures, it is a strong hint that the quadratic may not factorise neatly.

Example

Answers to 3 significant figures

Solve, giving your answers to 3 significant figures:

x^2 + y^2 = 18

2 x + y = 4

Rearrange the linear equation:
$y = 4 - 2 x$
Substitute into the quadratic equation:
$x^2 + (4 - 2x)^2 = 18$
Expand and simplify:
$x2+16−16x+4x2=185x2−16x−2=0\begin{aligned} x^2 + 16 - 16x + 4x^2 &= 18 \\ 5x^2 - 16x - 2 &= 0 \end{aligned}$
Use the quadratic formula with $a = 5$ , $b = - 16$ , and $c = - 2$ :
$\frac{16 \pm \sqrt{(-16)^2 - 4(5)(-2)}}{10}$
Simplify inside the square root:
$\frac{16 \pm \sqrt{296}}{10}$
Find the two $x$ -values:
$\approx 3.320 \quad \text{or} \quad x \approx -0.120$
Substitute into $y = 4 - 2 x$ :
$x≈−0.120,y≈4.241\begin{aligned} \text{If } x &\approx 3.320, & y &\approx -2.641 \\ \text{If } x &\approx -0.120, & y &\approx 4.241 \end{aligned}$
Round both coordinate pairs to 3 significant figures:
$\quad \text{and} \quad (-0.120, 4.24)$

Tip

Rounding decimals less than 1

For a number like -0.120, the zero after the 2 is significant. Writing -0.12 gives only 2 significant figures, so keep -0.120 when 3 significant figures are required.

Quadratics that are not circles

Not every quadratic simultaneous equation looks like $x^2 + y^2 = k$ . You may see equations such as $x^2 - 3y^2 = 13$ or $2x^2 - y^2 = 14$ .

The method is the same: use the linear equation to replace one variable, then solve the resulting quadratic.

Definition

Solve algebraically

To solve algebraically means to use algebraic methods such as rearranging, substituting, expanding and factorising. You should not rely on a graph for your final answer.

Example

Solving a quadratic with different squared terms

Solve algebraically:

2x^2 - y^2 = 1

3 x + 2 y = 1

Rearrange the linear equation to make $y$ the subject:
$2 y = 1 - 3 x$
Divide by 2:
$\frac{1 - 3x}{2}$
Substitute into $2x^2 - y^2 = 1$ :
$2x2−(1−3x2)2=12x^2 - \left(\frac{1 - 3x}{2}\right)^2 = 1$
Multiply every term by 4 to remove the fraction:
$8x^2 - (1 - 3x)^2 = 4$
Expand the bracket carefully:
$8x^2 - (1 - 6x + 9x^2) = 4$
Remove the bracket and simplify:
$x^2 + 6x - 1 = 4$
Rearrange into standard quadratic form:
$x^2 - 6x + 5 = 0$
Factorise:
$(x - 1) (x - 5) = 0$
Solve for $x$ :
$\quad \text{or} \quad x = 5$
Substitute into $\frac{1 - 3x}{2}$ :
$x=5,y=−7\begin{aligned} \text{If } x &= 1, & y &= -1 \\ \text{If } x &= 5, & y &= -7 \end{aligned}$
Write the two solutions:
$\quad \text{and} \quad (5, -7)$

Common Mistake

Forgetting the minus before a bracket

In expressions like $2x^2 - (1 - 3x)^2$ , the minus sign applies to the whole expanded bracket. Write the brackets out fully before simplifying.

Common Mistake

Sometimes there are not two real solutions

A line might touch the curve once or miss it completely. Algebraically, this shows up as a repeated root or a negative value under the square root in the quadratic formula.

Exam technique

In the exam

Rearrange the linear equation first, then substitute into the quadratic equation.
Keep your working neat when expanding brackets, especially when there is a minus sign before a squared bracket.
Always give answers as coordinate pairs and check each pair in both original equations if you have time.

Self review

Check yourself

Can you explain why a quadratic simultaneous equation can have two coordinate-pair solutions?
When substituting $y = 3 - 2 x$ into $x^2 + y^2 = 20$ , where do the brackets need to go?
If your quadratic gives two $x$ -values, how do you find the matching $y$ -values?

Quadratic Simultaneous Equations

What you'll learn

The big picture: line meets curve

The main method: substitute, solve, pair up

Rearranging a line with coefficients

When the quadratic does not factorise nicely

Quadratics that are not circles

Recap questions

You've reached the end

Quadratic Simultaneous Equations

What you'll learn

The big picture: line meets curve

The main method: substitute, solve, pair up

Rearranging a line with coefficients

When the quadratic does not factorise nicely

Quadratics that are not circles

Recap questions

You've reached the end

Quadratic Simultaneous Equations

What you'll learn

The big picture: line meets curve

The main method: substitute, solve, pair up

Rearranging a line with coefficients

When the quadratic does not factorise nicely

Quadratics that are not circles

Recap questions

You've reached the end

Quadratic Simultaneous Equations Revision Guide

Quadratic Simultaneous Equations

What you'll learn

The big picture: line meets curve

The main method: substitute, solve, pair up

Rearranging a line with coefficients

When the quadratic does not factorise nicely

Quadratics that are not circles

Recap questions

You've reached the end

Quadratic Simultaneous Equations Revision Guide