1.5.4 Simultaneous equations

What you'll learn

What it means to solve two equations simultaneously.
How to use elimination and substitution for two linear equations.
How to solve a system containing one linear and one quadratic equation.
How the number of solutions relates to intersections on a graph.

Prerequisites

Before starting, you should be comfortable with:

expanding brackets;
collecting like terms;
rearranging equations;
factorising and solving quadratic equations;
substituting a value into an expression.

For example, solving $x^2-5x+6=0$ by factorising gives

(x - 2) (x - 3) = 0,

so $x = 2$ or $x = 3$ .

You will use these algebra skills while finding values that satisfy more than one equation.

What are simultaneous equations?

Simultaneous equations are two or more equations involving the same unknowns. Their solution must satisfy every equation at the same time.

For two equations in the variables $x$ and $y$ , a solution is normally written as an ordered pair $(x, y)$ .

Definition

Solution of simultaneous equations

A solution is a set of values that makes all the equations true when the values are substituted into them.

For two linear equations, each equation represents a straight line. The simultaneous solution is the point where the two lines intersect.

Example

Checking a simultaneous solution

Show that $(x, y) = (3, 4)$ solves

x+y=7,2x−y=2.\begin{aligned} x+y&=7,\\ 2x-y&=2. \end{aligned}

Substitute $x = 3$ and $y = 4$ into the first equation:
$3 + 4 = 7,$
so the first equation is satisfied.
Substitute the same values into the second equation:
$2 (3) - 4 = 2,$
so the second equation is also satisfied.
Since both equations are true, $(3, 4)$ is the simultaneous solution.

Solving by elimination

The elimination method combines the equations so that one variable disappears. You can then solve an equation involving only the remaining variable.

Definition

Elimination

To eliminate a variable means to add or subtract equations so that the terms containing that variable cancel.

When the coefficients already match

If one variable has equal coefficients with the same sign, subtract the equations. If the coefficients have opposite signs, add the equations.

Example

Eliminating a variable directly

Solve

2x+3y=19,2x−y=7.\begin{aligned} 2x+3y&=19,\\ 2x-y&=7. \end{aligned}

The coefficients of $x$ are both $2$ , so subtract the second equation from the first:
$(2 x + 3 y) - (2 x - y) = 19 - 7.$
This gives
$4 y = 12,$
so $y = 3$ .
Substitute $y = 3$ into $2 x - y = 7$ :
$2 x - 3 = 7,$
giving $2 x = 10$ and therefore $x = 5$ .
Check the values in the other equation:
$2 (5) + 3 (3) = 10 + 9 = 19.$
Therefore, the solution is $(x, y) = (5, 3)$ .

Common Mistake

Subtracting negative terms

When subtracting an entire equation, every term changes sign. For example, $3 y - (- y) = 4 y$ , not $2 y$ .

When the coefficients do not match

You may need to multiply one or both equations by suitable numbers before adding or subtracting them.

Example

Creating matching coefficients

Solve

3x+2y=16,2x−3y=−5.\begin{aligned} 3x+2y&=16,\\ 2x-3y&=-5. \end{aligned}

To eliminate $y$ , create coefficients of $6 y$ . Multiply the first equation by 3 and the second by 2:
$9x+6y=48,4x−6y=−10.\begin{aligned} 9x+6y&=48,\\ 4x-6y&=-10. \end{aligned}$
Add the equations because $6 y$ and $- 6 y$ have opposite signs:
$13 x = 38,$
so
$x=3813.x=\frac{38}{13}.$
Substitute into $3 x + 2 y = 16$ :
$3(3813)+2y=16.3\left(\frac{38}{13}\right)+2y=16.$
Hence
$2y=20813−11413=9413,\begin{aligned} 2y&=\frac{208}{13}-\frac{114}{13}\\ &=\frac{94}{13}, \end{aligned}$
so
$y=4713.y=\frac{47}{13}.$
The solution is
$(x,y)=(3813,4713).(x,y)=\left(\frac{38}{13},\frac{47}{13}\right).$

Tip

Choosing what to eliminate

Look for the pair of coefficients whose lowest common multiple is easiest to create. This usually reduces the amount of arithmetic.

Solving by substitution

In the substitution method, you rearrange one equation to make one variable the subject, then replace that variable in the other equation.

Definition

Substitution

Substitution means replacing a variable with an equivalent expression or value.

This method is particularly convenient if one equation already has the form $y=…y=\ldots$ or $x=…x=\ldots$ .

Example

Substituting a linear expression

Solve

y=2x+1,3x+y=16.\begin{aligned} y&=2x+1,\\ 3x+y&=16. \end{aligned}

Since $y = 2 x + 1$ , substitute $2 x + 1$ for $y$ in the second equation:
$3 x + (2 x + 1) = 16.$
Solve the resulting equation:
$5x+1=16,5x=15,\begin{aligned} 5x+1&=16,\\ 5x&=15, \end{aligned}$
so $x = 3$ .
Use $y = 2 x + 1$ to find $y$ :
$y = 2 (3) + 1 = 7.$
Therefore, the solution is $(x, y) = (3, 7)$ .

Common Mistake

Substituting without brackets

When replacing a variable, use brackets around the substituted expression. For example, if $y = 2 x - 3$ , then $- 4 y = - 4 (2 x - 3)$ .

One linear and one quadratic equation

A linear equation contains variables only to the first power and represents a straight line.

A quadratic equation contains a squared variable, such as $x^2$ , and may represent a parabola.

To solve one linear and one quadratic equation simultaneously:

Rearrange the linear equation to make one variable the subject.
Substitute this expression into the quadratic equation.
Solve the resulting quadratic.
Find the corresponding value of the other variable for each solution.

Key Idea

Expecting two solutions

A straight line can intersect a parabola twice, so a linear-quadratic system will often produce two ordered pairs. Do not stop after finding the first value.

Example

Solving a linear-quadratic system

Solve

y=x+1,y=x2−3x+1.\begin{aligned} y&=x+1,\\ y&=x^2-3x+1. \end{aligned}

Both expressions equal $y$ , so set them equal:
$x+1=x^2-3x+1.$
Rearrange to form a quadratic equation:
$0=x2−4x,0=x(x−4).\begin{aligned} 0&=x^2-4x,\\ 0&=x(x-4). \end{aligned}$
Therefore, $x = 0$ or $x = 4$ .
Substitute each value into $y = x + 1$ :
$x=0⇒y=1,x=4⇒y=5.\begin{aligned} x=0&\Rightarrow y=1,\\ x=4&\Rightarrow y=5. \end{aligned}$
The two solutions are
$(x,y)=(0,1)or(x,y)=(4,5).(x,y)=(0,1)\quad\text{or}\quad(x,y)=(4,5).$

The solutions are exactly the intersection points of the line and the parabola.

Graph of the line y equals x plus 1 and the parabola y equals x squared minus 3x plus 1, intersecting at 0 comma 1 and 4 comma 5

The possible numbers of solutions

A line and a parabola can have:

two solutions if they intersect twice;
one repeated solution if the line is tangent to the parabola;
no real solutions if they do not intersect.

After substitution, these possibilities appear in the resulting quadratic equation. If it has the form $ax^2+bx+c=0$ , its discriminant is $b^2-4ac$ .

If $b^2-4ac>0$ , there are two distinct real solutions.
If $b^2-4ac=0$ , there is one repeated real solution.
If $b^2-4ac<0$ , there are no real solutions.

Common Mistake

No real solutions

A negative discriminant does not mean your algebra must be wrong. It can mean that the line and quadratic curve do not meet anywhere on a real coordinate graph.

Checking your answers

Substitute every ordered pair into both original equations. This catches sign errors and confirms that you have matched each $x$ -value with the correct $y$ -value.

Common Mistake

Mixing corresponding values

If there are two solutions, calculate the value of $y$ separately for each value of $x$ . Do not combine an $x$ -value from one solution with the $y$ -value from the other.

Exam technique

In the exam

Label equations if you are using elimination, and show clearly what each equation is multiplied by.
After substitution, rearrange into the standard quadratic form $ax^2+bx+c=0$ before factorising or using the quadratic formula.
Continue until you have complete ordered pairs, and check each pair in both original equations.
Keep answers exact unless the question asks for a decimal approximation.

Self review

Check yourself

How would you decide whether elimination or substitution is the more efficient method?
Can you solve $2 x + y = 8$ and $3 x - y = 7$ ?
How many real solutions are possible when solving one linear and one quadratic equation simultaneously?

What you'll learn

What it means to solve two equations simultaneously.
How to use elimination and substitution for two linear equations.
How to solve a system containing one linear and one quadratic equation.
How the number of solutions relates to intersections on a graph.

Prerequisites

Before starting, you should be comfortable with:

expanding brackets;
collecting like terms;
rearranging equations;
factorising and solving quadratic equations;
substituting a value into an expression.

For example, solving $x^2-5x+6=0$ by factorising gives

(x - 2) (x - 3) = 0,

so $x = 2$ or $x = 3$ .

You will use these algebra skills while finding values that satisfy more than one equation.

What are simultaneous equations?

Simultaneous equations are two or more equations involving the same unknowns. Their solution must satisfy every equation at the same time.

For two equations in the variables $x$ and $y$ , a solution is normally written as an ordered pair $(x, y)$ .

Definition

Solution of simultaneous equations

A solution is a set of values that makes all the equations true when the values are substituted into them.

For two linear equations, each equation represents a straight line. The simultaneous solution is the point where the two lines intersect.

Example

Checking a simultaneous solution

Show that $(x, y) = (3, 4)$ solves

x+y=7,2x−y=2.\begin{aligned} x+y&=7,\\ 2x-y&=2. \end{aligned}

Substitute $x = 3$ and $y = 4$ into the first equation:
$3 + 4 = 7,$
so the first equation is satisfied.
Substitute the same values into the second equation:
$2 (3) - 4 = 2,$
so the second equation is also satisfied.
Since both equations are true, $(3, 4)$ is the simultaneous solution.

Solving by elimination

The elimination method combines the equations so that one variable disappears. You can then solve an equation involving only the remaining variable.

Definition

Elimination

To eliminate a variable means to add or subtract equations so that the terms containing that variable cancel.

When the coefficients already match

If one variable has equal coefficients with the same sign, subtract the equations. If the coefficients have opposite signs, add the equations.

Example

Eliminating a variable directly

Solve

2x+3y=19,2x−y=7.\begin{aligned} 2x+3y&=19,\\ 2x-y&=7. \end{aligned}

The coefficients of $x$ are both $2$ , so subtract the second equation from the first:
$(2 x + 3 y) - (2 x - y) = 19 - 7.$
This gives
$4 y = 12,$
so $y = 3$ .
Substitute $y = 3$ into $2 x - y = 7$ :
$2 x - 3 = 7,$
giving $2 x = 10$ and therefore $x = 5$ .
Check the values in the other equation:
$2 (5) + 3 (3) = 10 + 9 = 19.$
Therefore, the solution is $(x, y) = (5, 3)$ .

Common Mistake

Subtracting negative terms

When subtracting an entire equation, every term changes sign. For example, $3 y - (- y) = 4 y$ , not $2 y$ .

When the coefficients do not match

You may need to multiply one or both equations by suitable numbers before adding or subtracting them.

Example

Creating matching coefficients

Solve

3x+2y=16,2x−3y=−5.\begin{aligned} 3x+2y&=16,\\ 2x-3y&=-5. \end{aligned}

To eliminate $y$ , create coefficients of $6 y$ . Multiply the first equation by 3 and the second by 2:
$9x+6y=48,4x−6y=−10.\begin{aligned} 9x+6y&=48,\\ 4x-6y&=-10. \end{aligned}$
Add the equations because $6 y$ and $- 6 y$ have opposite signs:
$13 x = 38,$
so
$x=3813.x=\frac{38}{13}.$
Substitute into $3 x + 2 y = 16$ :
$3(3813)+2y=16.3\left(\frac{38}{13}\right)+2y=16.$
Hence
$2y=20813−11413=9413,\begin{aligned} 2y&=\frac{208}{13}-\frac{114}{13}\\ &=\frac{94}{13}, \end{aligned}$
so
$y=4713.y=\frac{47}{13}.$
The solution is
$(x,y)=(3813,4713).(x,y)=\left(\frac{38}{13},\frac{47}{13}\right).$

Tip

Choosing what to eliminate

Look for the pair of coefficients whose lowest common multiple is easiest to create. This usually reduces the amount of arithmetic.

Solving by substitution

In the substitution method, you rearrange one equation to make one variable the subject, then replace that variable in the other equation.

Definition

Substitution

Substitution means replacing a variable with an equivalent expression or value.

This method is particularly convenient if one equation already has the form $y=…y=\ldots$ or $x=…x=\ldots$ .

Example

Substituting a linear expression

Solve

y=2x+1,3x+y=16.\begin{aligned} y&=2x+1,\\ 3x+y&=16. \end{aligned}

Since $y = 2 x + 1$ , substitute $2 x + 1$ for $y$ in the second equation:
$3 x + (2 x + 1) = 16.$
Solve the resulting equation:
$5x+1=16,5x=15,\begin{aligned} 5x+1&=16,\\ 5x&=15, \end{aligned}$
so $x = 3$ .
Use $y = 2 x + 1$ to find $y$ :
$y = 2 (3) + 1 = 7.$
Therefore, the solution is $(x, y) = (3, 7)$ .

Common Mistake

Substituting without brackets

When replacing a variable, use brackets around the substituted expression. For example, if $y = 2 x - 3$ , then $- 4 y = - 4 (2 x - 3)$ .

One linear and one quadratic equation

A linear equation contains variables only to the first power and represents a straight line.

A quadratic equation contains a squared variable, such as $x^2$ , and may represent a parabola.

To solve one linear and one quadratic equation simultaneously:

Rearrange the linear equation to make one variable the subject.
Substitute this expression into the quadratic equation.
Solve the resulting quadratic.
Find the corresponding value of the other variable for each solution.

Key Idea

Expecting two solutions

A straight line can intersect a parabola twice, so a linear-quadratic system will often produce two ordered pairs. Do not stop after finding the first value.

Example

Solving a linear-quadratic system

Solve

y=x+1,y=x2−3x+1.\begin{aligned} y&=x+1,\\ y&=x^2-3x+1. \end{aligned}

Both expressions equal $y$ , so set them equal:
$x+1=x^2-3x+1.$
Rearrange to form a quadratic equation:
$0=x2−4x,0=x(x−4).\begin{aligned} 0&=x^2-4x,\\ 0&=x(x-4). \end{aligned}$
Therefore, $x = 0$ or $x = 4$ .
Substitute each value into $y = x + 1$ :
$x=0⇒y=1,x=4⇒y=5.\begin{aligned} x=0&\Rightarrow y=1,\\ x=4&\Rightarrow y=5. \end{aligned}$
The two solutions are
$(x,y)=(0,1)or(x,y)=(4,5).(x,y)=(0,1)\quad\text{or}\quad(x,y)=(4,5).$

The solutions are exactly the intersection points of the line and the parabola.

Graph of the line y equals x plus 1 and the parabola y equals x squared minus 3x plus 1, intersecting at 0 comma 1 and 4 comma 5

The possible numbers of solutions

A line and a parabola can have:

two solutions if they intersect twice;
one repeated solution if the line is tangent to the parabola;
no real solutions if they do not intersect.

After substitution, these possibilities appear in the resulting quadratic equation. If it has the form $ax^2+bx+c=0$ , its discriminant is $b^2-4ac$ .

If $b^2-4ac>0$ , there are two distinct real solutions.
If $b^2-4ac=0$ , there is one repeated real solution.
If $b^2-4ac<0$ , there are no real solutions.

Common Mistake

No real solutions

A negative discriminant does not mean your algebra must be wrong. It can mean that the line and quadratic curve do not meet anywhere on a real coordinate graph.

Checking your answers

Substitute every ordered pair into both original equations. This catches sign errors and confirms that you have matched each $x$ -value with the correct $y$ -value.

Common Mistake

Mixing corresponding values

If there are two solutions, calculate the value of $y$ separately for each value of $x$ . Do not combine an $x$ -value from one solution with the $y$ -value from the other.

Exam technique

In the exam

Label equations if you are using elimination, and show clearly what each equation is multiplied by.
After substitution, rearrange into the standard quadratic form $ax^2+bx+c=0$ before factorising or using the quadratic formula.
Continue until you have complete ordered pairs, and check each pair in both original equations.
Keep answers exact unless the question asks for a decimal approximation.

Self review

Check yourself

How would you decide whether elimination or substitution is the more efficient method?
Can you solve $2 x + y = 8$ and $3 x - y = 7$ ?
How many real solutions are possible when solving one linear and one quadratic equation simultaneously?

What you'll learn

Prerequisites

What are simultaneous equations?

Solving by elimination

When the coefficients already match

When the coefficients do not match

Solving by substitution

One linear and one quadratic equation

The possible numbers of solutions

Checking your answers

Review 1.5.4 Simultaneous equations by teaching Genie

1.5.4 Simultaneous equations

What you'll learn

Prerequisites

What are simultaneous equations?

Solving by elimination

When the coefficients already match

When the coefficients do not match

Solving by substitution

One linear and one quadratic equation

The possible numbers of solutions

Checking your answers

Review 1.5.4 Simultaneous equations by teaching Genie

1.5.4 Simultaneous equations

What you'll learn

Prerequisites

What are simultaneous equations?

Solving by elimination

When the coefficients already match

When the coefficients do not match

Solving by substitution

One linear and one quadratic equation

The possible numbers of solutions

Checking your answers

Review 1.5.4 Simultaneous equations by teaching Genie

1.5.4 Simultaneous equations Revision Guide

1.5.4 Simultaneous equations

What you'll learn

Prerequisites

What are simultaneous equations?

Solving by elimination

When the coefficients already match

When the coefficients do not match

Solving by substitution

One linear and one quadratic equation

The possible numbers of solutions

Checking your answers

Review 1.5.4 Simultaneous equations by teaching Genie

1.5.4 Simultaneous equations Revision Guide