Quadratic Simultaneous Equations

What you'll learn

How to recognise a linear equation paired with a quadratic equation.
How substitution turns the pair into one quadratic in one variable.
How to solve by factorising or using the quadratic formula.
How to keep answer pairs matched, including when rounding to 3 significant figures.

1. The big picture

Definition

Simultaneous equations

Simultaneous equations are equations that must be true at the same time. A solution is an ordered pair, such as $(x, y)$ , that satisfies every equation in the set.

An ordered pair lists the $x$ -value first and the $y$ -value second. For example, $(2, 5)$ means $x = 2$ and $y = 5$ .

A linear equation has variables only to the first power, such as $2 x + y = 7$ ; its graph is a straight line. A quadratic equation has a squared term, such as $x^2 + y^2 = 34$ or $2x^2 - y^2 = 14$ ; its graph is curved.

A linear graph is a straight line, while a quadratic relation gives a curved graph; solutions are where the graphs meet.

Example

Checking a possible solution

Suppose you are checking whether $(2, 5)$ solves these equations:

The point (2, 5) lies on both the line and the circle, so it satisfies both equations.

$$
\begin{aligned}
x^2 + y^2 &= 29 \\
y &= x + 3
\end{aligned}
$$

2. Test the linear equation $y = x + 3$ .

$$
5 = 2 + 3
$$

3. Test the quadratic equation $x^2 + y^2 = 29$ .

$$
2^2 + 5^2 = 4 + 25 = 29
$$

4. Both equations are true, so $(2, 5)$ is a solution.

2. The main method: substitution

Substitution means replacing a variable with an equal expression. The subject of an equation is the variable isolated on one side; in $x = y - 1$ , $x$ is the subject.

Key Idea

The method to remember

Use the linear equation to replace one variable in the quadratic equation. This gives you one quadratic equation in one variable.

Definition

Standard form of a quadratic

A quadratic in one variable is in standard form when it is written as $ax^2 + bx + c = 0$ , where $\neq 0$ .

Factorising means rewriting an expression as a product of brackets. The roots of an equation are its solutions.

Example

Substitution when a variable is already the subject

Solve the simultaneous equations:

The solutions are the intersection points of the circle x^2+y^2=25 and the line x=y-1.

$$
\begin{aligned}
x^2 + y^2 &= 25 \\
x &= y - 1
\end{aligned}
$$

2. Substitute $x = y - 1$ into $x^2 + y^2 = 25$ .

$$
(y - 1)^2 + y^2 = 25
$$

3. Expand the bracket.

$$
y^2 - 2y + 1 + y^2 = 25
$$

4. Rearrange into standard form, then divide by 2.

$$
\begin{aligned}
2y^2 - 2y - 24 &= 0 \\
y^2 - y - 12 &= 0
\end{aligned}
$$

5. Factorise.

$$
(y - 4)(y + 3) = 0
$$

6. Therefore $y = 4$ or $y = - 3$ .

Use $x = y - 1$ to find the matching $x$ -values: if $y = 4$ , then $x = 3$ ; if $y = - 3$ , then $x = - 4$ .
The solutions are $(3, 4)$ and $(- 4, - 3)$ .

Common Mistake

Unpaired answers

Writing separate lists of $x$ -values and $y$ -values is not enough unless the pairings are clear. Always give coordinate pairs.

3. When the linear equation needs rearranging

Sometimes the linear equation is not already written as $\dots$ or $\dots$ . Rearrange it first, choosing the variable that looks easiest to substitute.

Example

Rearrange the line first

Solve:

Rearranging the line first still gives the same visual idea: find where the line cuts the circle.

$$
\begin{aligned}
x^2 + y^2 &= 25 \\
3x + y &= 5
\end{aligned}
$$

2. Make $y$ the subject of the linear equation.

$$
y = 5 - 3x
$$

3. Substitute into the quadratic equation and simplify.

$$
\begin{aligned}
x^2 + (5 - 3x)^2 &= 25 \\
x^2 + 25 - 30x + 9x^2 &= 25 \\
10x^2 - 30x &= 0
\end{aligned}
$$

4. Factorise and solve.

$$
10x(x - 3) = 0
$$

5. So $x = 0$ or $x = 3$ .

Use $y = 5 - 3 x$ : if $x = 0$ , then $y = 5$ ; if $x = 3$ , then $y = - 4$ .
The solutions are $(0, 5)$ and $(3, - 4)$ .

Common Mistake

Squaring a whole expression

When substituting $y = 5 - 3 x$ , write $5 - 3x)^2$ . Do not treat it as $5 - 3x^2$ .

4. When the quadratic does not factorise nicely

If factorising is awkward, use the quadratic formula. A coefficient is the number multiplying a term; in $5x^2 - 16x - 2 = 0$ , the coefficients are $a = 5$ , $b = - 16$ and $c = - 2$ .

For $ax^2 + bx + c = 0$ :

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

The expression $b^2 - 4ac$ is the discriminant. If it is positive, you get two real roots; if it is zero, you get one repeated root.

Common Mistake

When there are no real answers

If the discriminant $b^2 - 4ac$ is negative, the square root is not a real number. That means the line does not meet the curve in real coordinate pairs.

A negative discriminant means the line misses the curve, so there are no real intersection points.

Example

Answers to 3 significant figures

Solve, giving answers to 3 significant figures:

When the intersections are not neat, the coordinates can be found using the quadratic formula and rounded at the end.

$$
\begin{aligned}
x^2 + y^2 &= 18 \\
2x + y &= 4
\end{aligned}
$$

2. Make $y$ the subject.

$$
y = 4 - 2x
$$

3. Substitute and collect terms.

$$
\begin{aligned}
x^2 + (4 - 2x)^2 &= 18 \\
x^2 + 16 - 16x + 4x^2 &= 18 \\
5x^2 - 16x - 2 &= 0
\end{aligned}
$$

4. Use the quadratic formula.

$$
x = \frac{16 \pm \sqrt{296}}{10}
$$

5. Calculator values give $\approx 3.320\ldots$ or $\approx -0.120\ldots$ .

Use $y = 4 - 2 x$ with the unrounded values: $\approx -2.640\ldots$ or $\approx 4.240\ldots$ .
To 3 significant figures, the solutions are $(3.32, - 2.64)$ and $(- 0.120, 4.24)$ .

Tip

Rounding safely

Keep full calculator values until the final line. Rounding too early can make the matching coordinate slightly inaccurate.

5. Harder algebraic examples

Definition

Solve algebraically

Solving algebraically means using equation steps such as rearranging, substituting, expanding and factorising, rather than reading answers from a graph or table.

Some Grade 8/9 questions include equations such as $x^2 - 2y^2 = 7$ or $2x^2 - y^2 = 14$ . The method is still the same: rearrange the line, substitute, then solve the quadratic.

Example

A quadratic with different squared terms

Solve algebraically:

The same substitution method works when the quadratic curve is not a circle.

$$
\begin{aligned}
x^2 - 2y^2 &= 7 \\
x + 2y &= 5
\end{aligned}
$$

2. Make $x$ the subject of the linear equation.

$$
x = 5 - 2y
$$

3. Substitute into the quadratic equation.

$$
(5 - 2y)^2 - 2y^2 = 7
$$

4. Expand and simplify.

$$
\begin{aligned}
25 - 20y + 4y^2 - 2y^2 &= 7 \\
2y^2 - 20y + 18 &= 0
\end{aligned}
$$

5. Divide by 2.

$$
y^2 - 10y + 9 = 0
$$

6. Factorise.

$$
(y - 1)(y - 9) = 0
$$

7. So $y = 1$ or $y = 9$ .

Use $x = 5 - 2 y$ : if $y = 1$ , then $x = 3$ ; if $y = 9$ , then $x = - 13$ .
The solutions are $(3, 1)$ and $(- 13, 9)$ .

Exam technique

In the exam

Rearrange the linear equation to make the simpler variable the subject.
Substitute carefully using brackets, then expand and collect into standard form.
Solve the quadratic, pair each root with its matching value, and round only at the end if asked.

Self review

Check yourself

Can you explain why a quadratic simultaneous equation can have two solution pairs?
When substituting $y = 3 - 2 x$ , where must the brackets go?
If a question asks for answers to 3 significant figures, when should you round?

What you'll learn

How to recognise a linear equation paired with a quadratic equation.
How substitution turns the pair into one quadratic in one variable.
How to solve by factorising or using the quadratic formula.
How to keep answer pairs matched, including when rounding to 3 significant figures.

1. The big picture

Definition

Simultaneous equations

Simultaneous equations are equations that must be true at the same time. A solution is an ordered pair, such as $(x, y)$ , that satisfies every equation in the set.

An ordered pair lists the $x$ -value first and the $y$ -value second. For example, $(2, 5)$ means $x = 2$ and $y = 5$ .

A linear graph is a straight line, while a quadratic relation gives a curved graph; solutions are where the graphs meet.

Example

Checking a possible solution

Suppose you are checking whether $(2, 5)$ solves these equations:

The point (2, 5) lies on both the line and the circle, so it satisfies both equations.

$$
\begin{aligned}
x^2 + y^2 &= 29 \\
y &= x + 3
\end{aligned}
$$

2. Test the linear equation $y = x + 3$ .

$$
5 = 2 + 3
$$

3. Test the quadratic equation $x^2 + y^2 = 29$ .

$$
2^2 + 5^2 = 4 + 25 = 29
$$

4. Both equations are true, so $(2, 5)$ is a solution.

2. The main method: substitution

Substitution means replacing a variable with an equal expression. The subject of an equation is the variable isolated on one side; in $x = y - 1$ , $x$ is the subject.

Key Idea

The method to remember

Use the linear equation to replace one variable in the quadratic equation. This gives you one quadratic equation in one variable.

Definition

Standard form of a quadratic

A quadratic in one variable is in standard form when it is written as $ax^2 + bx + c = 0$ , where $\neq 0$ .

Factorising means rewriting an expression as a product of brackets. The roots of an equation are its solutions.

Example

Substitution when a variable is already the subject

Solve the simultaneous equations:

The solutions are the intersection points of the circle x^2+y^2=25 and the line x=y-1.

$$
\begin{aligned}
x^2 + y^2 &= 25 \\
x &= y - 1
\end{aligned}
$$

2. Substitute $x = y - 1$ into $x^2 + y^2 = 25$ .

$$
(y - 1)^2 + y^2 = 25
$$

3. Expand the bracket.

$$
y^2 - 2y + 1 + y^2 = 25
$$

4. Rearrange into standard form, then divide by 2.

$$
\begin{aligned}
2y^2 - 2y - 24 &= 0 \\
y^2 - y - 12 &= 0
\end{aligned}
$$

5. Factorise.

$$
(y - 4)(y + 3) = 0
$$

6. Therefore $y = 4$ or $y = - 3$ .

Use $x = y - 1$ to find the matching $x$ -values: if $y = 4$ , then $x = 3$ ; if $y = - 3$ , then $x = - 4$ .
The solutions are $(3, 4)$ and $(- 4, - 3)$ .

Common Mistake

Unpaired answers

Writing separate lists of $x$ -values and $y$ -values is not enough unless the pairings are clear. Always give coordinate pairs.

3. When the linear equation needs rearranging

Sometimes the linear equation is not already written as $\dots$ or $\dots$ . Rearrange it first, choosing the variable that looks easiest to substitute.

Example

Rearrange the line first

Solve:

Rearranging the line first still gives the same visual idea: find where the line cuts the circle.

$$
\begin{aligned}
x^2 + y^2 &= 25 \\
3x + y &= 5
\end{aligned}
$$

2. Make $y$ the subject of the linear equation.

$$
y = 5 - 3x
$$

3. Substitute into the quadratic equation and simplify.

$$
\begin{aligned}
x^2 + (5 - 3x)^2 &= 25 \\
x^2 + 25 - 30x + 9x^2 &= 25 \\
10x^2 - 30x &= 0
\end{aligned}
$$

4. Factorise and solve.

$$
10x(x - 3) = 0
$$

5. So $x = 0$ or $x = 3$ .

Use $y = 5 - 3 x$ : if $x = 0$ , then $y = 5$ ; if $x = 3$ , then $y = - 4$ .
The solutions are $(0, 5)$ and $(3, - 4)$ .

Common Mistake

Squaring a whole expression

When substituting $y = 5 - 3 x$ , write $5 - 3x)^2$ . Do not treat it as $5 - 3x^2$ .

4. When the quadratic does not factorise nicely

If factorising is awkward, use the quadratic formula. A coefficient is the number multiplying a term; in $5x^2 - 16x - 2 = 0$ , the coefficients are $a = 5$ , $b = - 16$ and $c = - 2$ .

For $ax^2 + bx + c = 0$ :

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

The expression $b^2 - 4ac$ is the discriminant. If it is positive, you get two real roots; if it is zero, you get one repeated root.

Common Mistake

When there are no real answers

If the discriminant $b^2 - 4ac$ is negative, the square root is not a real number. That means the line does not meet the curve in real coordinate pairs.

A negative discriminant means the line misses the curve, so there are no real intersection points.

Example

Answers to 3 significant figures

Solve, giving answers to 3 significant figures:

When the intersections are not neat, the coordinates can be found using the quadratic formula and rounded at the end.

$$
\begin{aligned}
x^2 + y^2 &= 18 \\
2x + y &= 4
\end{aligned}
$$

2. Make $y$ the subject.

$$
y = 4 - 2x
$$

3. Substitute and collect terms.

$$
\begin{aligned}
x^2 + (4 - 2x)^2 &= 18 \\
x^2 + 16 - 16x + 4x^2 &= 18 \\
5x^2 - 16x - 2 &= 0
\end{aligned}
$$

4. Use the quadratic formula.

$$
x = \frac{16 \pm \sqrt{296}}{10}
$$

5. Calculator values give $\approx 3.320\ldots$ or $\approx -0.120\ldots$ .

Use $y = 4 - 2 x$ with the unrounded values: $\approx -2.640\ldots$ or $\approx 4.240\ldots$ .
To 3 significant figures, the solutions are $(3.32, - 2.64)$ and $(- 0.120, 4.24)$ .

Tip

Rounding safely

Keep full calculator values until the final line. Rounding too early can make the matching coordinate slightly inaccurate.

5. Harder algebraic examples

Definition

Solve algebraically

Solving algebraically means using equation steps such as rearranging, substituting, expanding and factorising, rather than reading answers from a graph or table.

Some Grade 8/9 questions include equations such as $x^2 - 2y^2 = 7$ or $2x^2 - y^2 = 14$ . The method is still the same: rearrange the line, substitute, then solve the quadratic.

Example

A quadratic with different squared terms

Solve algebraically:

The same substitution method works when the quadratic curve is not a circle.

$$
\begin{aligned}
x^2 - 2y^2 &= 7 \\
x + 2y &= 5
\end{aligned}
$$

2. Make $x$ the subject of the linear equation.

$$
x = 5 - 2y
$$

3. Substitute into the quadratic equation.

$$
(5 - 2y)^2 - 2y^2 = 7
$$

4. Expand and simplify.

$$
\begin{aligned}
25 - 20y + 4y^2 - 2y^2 &= 7 \\
2y^2 - 20y + 18 &= 0
\end{aligned}
$$

5. Divide by 2.

$$
y^2 - 10y + 9 = 0
$$

6. Factorise.

$$
(y - 1)(y - 9) = 0
$$

7. So $y = 1$ or $y = 9$ .

Use $x = 5 - 2 y$ : if $y = 1$ , then $x = 3$ ; if $y = 9$ , then $x = - 13$ .
The solutions are $(3, 1)$ and $(- 13, 9)$ .

Exam technique

In the exam

Rearrange the linear equation to make the simpler variable the subject.
Substitute carefully using brackets, then expand and collect into standard form.
Solve the quadratic, pair each root with its matching value, and round only at the end if asked.

Self review

Check yourself

Can you explain why a quadratic simultaneous equation can have two solution pairs?
When substituting $y = 3 - 2 x$ , where must the brackets go?
If a question asks for answers to 3 significant figures, when should you round?

What you'll learn

1. The big picture

2. The main method: substitution

3. When the linear equation needs rearranging

4. When the quadratic does not factorise nicely

5. Harder algebraic examples

Recap questions

Review Quadratic Simultaneous Equations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Simultaneous Equations

What you'll learn

1. The big picture

2. The main method: substitution

3. When the linear equation needs rearranging

4. When the quadratic does not factorise nicely

5. Harder algebraic examples

Recap questions

Review Quadratic Simultaneous Equations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Simultaneous Equations

What you'll learn

1. The big picture

2. The main method: substitution

3. When the linear equation needs rearranging

4. When the quadratic does not factorise nicely

5. Harder algebraic examples

Recap questions

Review Quadratic Simultaneous Equations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Simultaneous Equations Revision Guide

Quadratic Simultaneous Equations

What you'll learn

1. The big picture

2. The main method: substitution

3. When the linear equation needs rearranging

4. When the quadratic does not factorise nicely

5. Harder algebraic examples

Recap questions

Review Quadratic Simultaneous Equations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Quadratic Simultaneous Equations Revision Guide