Simultaneous Equations

What you'll learn

What simultaneous equations are and what a solution means.
How to solve linear simultaneous equations by elimination.
How to handle rearranged equations and decimals.
How to turn short word problems into equations.

The basic idea

A linear equation is an equation where the variables only have power 1, such as $4 x + 3 y = 18$ . Its graph is a straight line.

Definition

Simultaneous equations

Simultaneous equations are two or more equations that are true at the same time. For IGCSE Grade 5, you usually solve two linear equations with two variables, often $x$ and $y$ .

A solution is the pair of values that makes both equations true. For example, $x = 2$ and $y = 3$ is written as the solution pair $(2, 3)$ .

Graphically, each linear equation is a straight line. The simultaneous solution is where the two lines cross.

Two straight-line graphs crossing at the labelled solution point (2, 3)

Key Idea

The big picture

Solving simultaneous equations means finding the one value of $x$ and the one value of $y$ that work in both equations.

Example

Checking whether a pair is a solution

Check whether $x = 4$ and $y = 1$ solves the equations $2 x + 3 y = 11$ and $5 x - y = 19$ .

Substitute $x = 4$ and $y = 1$ into the first equation.
$2 (4) + 3 (1) = 8 + 3 = 11$
The first equation works because the left side equals 11.
Substitute $x = 4$ and $y = 1$ into the second equation.
$5 (4) - 1 = 20 - 1 = 19$
The second equation also works, so $x = 4$ and $y = 1$ is the solution.

The elimination method

The main Grade 5 method is elimination.

Definition

Eliminate

To eliminate a variable means to remove it from the equations by adding or subtracting the equations.

You try to make either the $x$ terms or the $y$ terms cancel out. Then you solve for the remaining variable.

When the signs are opposite: add

If one equation has $+ 3 y$ and the other has $- 3 y$ , adding the equations cancels the $y$ terms.

Example

Eliminating by adding

Solve the simultaneous equations $4 x + 3 y = 26$ and $x - 3 y = - 1$ .

Write the equations clearly and label them.
$4x+3y=26(1)x−3y=−1(2)\begin{aligned} 4x+3y&=26 \quad \text{(1)}\\ x-3y&=-1 \quad \text{(2)} \end{aligned}$
The $y$ terms are opposites, $+ 3 y$ and $- 3 y$ , so add the two equations.
$4x+x=5x3y−3y=026+(−1)=25\begin{aligned} 4x+x&=5x\\ 3y-3y&=0\\ 26+(-1)&=25 \end{aligned}$
Solve the new equation.
$5 x = 25$
Divide by 5.
$x = 5$
Substitute $x = 5$ into equation (2).
$5 - 3 y = - 1$
Solve for $y$ .
$−3y=−6y=2\begin{aligned} -3y&=-6\\ y&=2 \end{aligned}$
Give both values.
$x=5,y=2x=5,\quad y=2$

Tip

Add or subtract?

If the matching terms have opposite signs, add the equations. If they have the same sign, subtract the equations.

When the signs are the same: subtract

If both equations have the same $x$ coefficient, such as $2 x$ and $2 x$ , subtracting can eliminate $x$ .

Example

Eliminating by subtracting

Solve the simultaneous equations $2 x + 5 y = 28$ and $2 x - y = 4$ .

Label the equations.
$2x+5y=28(1)2x−y=4(2)\begin{aligned} 2x+5y&=28 \quad \text{(1)}\\ 2x-y&=4 \quad \text{(2)} \end{aligned}$
The $x$ terms are the same, so subtract equation (2) from equation (1).
$(2x+5y)−(2x−y)=28−46y=24\begin{aligned} (2x+5y)-(2x-y)&=28-4\\ 6y&=24 \end{aligned}$
Solve for $y$ .
$y = 4$
Substitute $y = 4$ into equation (2).
$2 x - 4 = 4$
Solve for $x$ .
$2x=8x=4\begin{aligned} 2x&=8\\ x&=4 \end{aligned}$
State the solution.
$x=4,y=4x=4,\quad y=4$

Common Mistake

Subtracting negatives incorrectly

Be careful with signs. In the example above, subtracting $- y$ means $5 y - (- y) = 6 y$ , not $4 y$ .

Making a coefficient match

Sometimes neither variable cancels straight away. A coefficient is the number multiplying a variable, such as the 4 in $4 x$ .

To eliminate a variable, you can multiply one whole equation so that a coefficient matches.

Example

Multiplying first, then eliminating

Solve $2 x - 3 y = 4$ and $4 x - y = 18$ .

Label the equations.
$2x−3y=4(1)4x−y=18(2)\begin{aligned} 2x-3y&=4 \quad \text{(1)}\\ 4x-y&=18 \quad \text{(2)} \end{aligned}$
The first equation has $2 x$ and the second has $4 x$ . Multiply equation (1) by 2 so the $x$ coefficients match.
$\quad \text{(3)}$
Now subtract equation (2) from equation (3).
$(4x−6y)−(4x−y)=8−18−5y=−10\begin{aligned} (4x-6y)-(4x-y)&=8-18\\ -5y&=-10 \end{aligned}$
Solve for $y$ .
$y = 2$
Substitute $y = 2$ into equation (2).
$4 x - 2 = 18$
Solve for $x$ .
$4x=20x=5\begin{aligned} 4x&=20\\ x&=5 \end{aligned}$
Write the final answer.
$x=5,y=2x=5,\quad y=2$

Common Mistake

Only multiplying part of an equation

If you multiply an equation by 2, multiply every term on both sides. For example, $2 x - 3 y = 4$ becomes $4 x - 6 y = 8$ .

Rearranged equations

Not every question starts in the neat form $a x + b y = c$ . You may first need to rearrange.

Definition

Standard form

For this topic, standard form means writing a linear equation like $a x + b y = c$ , with the $x$ and $y$ terms on the left and the number on the right.

Example

Rearranging before solving

Solve $3 x = 10 + y$ and $x + 4 y = 12$ .

Rearrange the first equation by subtracting $y$ from both sides.
$3 x - y = 10$
Label the two equations.
$3x−y=10(1)x+4y=12(2)\begin{aligned} 3x-y&=10 \quad \text{(1)}\\ x+4y&=12 \quad \text{(2)} \end{aligned}$
Multiply equation (2) by 3 so the $x$ coefficients match.
$\quad \text{(3)}$
Subtract equation (1) from equation (3).
$(3x+12y)−(3x−y)=36−1013y=26\begin{aligned} (3x+12y)-(3x-y)&=36-10\\ 13y&=26 \end{aligned}$
Solve for $y$ .
$y = 2$
Substitute $y = 2$ into equation (2).
$x + 8 = 12$
Solve for $x$ .
$x = 4$
State the solution.
$x=4,y=2x=4,\quad y=2$

Decimals in simultaneous equations

Decimals do not change the method. You can either work with decimals carefully or multiply everything by 10 to clear one decimal place.

Tip

Clearing decimals

If an equation contains 4.5, multiplying the whole equation by 10 changes it to 45. This can make elimination easier.

Example

Solving with decimals

Solve $6 x + 5 y = 6.5$ and $3 x - 2 y = 5.5$ .

Label the equations.
$6x+5y=6.5(1)3x−2y=5.5(2)\begin{aligned} 6x+5y&=6.5 \quad \text{(1)}\\ 3x-2y&=5.5 \quad \text{(2)} \end{aligned}$
Multiply equation (2) by 2 so the $x$ coefficients match.
$\quad \text{(3)}$
Subtract equation (3) from equation (1).
$(6x+5y)−(6x−4y)=6.5−119y=−4.5\begin{aligned} (6x+5y)-(6x-4y)&=6.5-11\\ 9y&=-4.5 \end{aligned}$
Solve for $y$ .
$y = - 0.5$
Substitute $y = - 0.5$ into equation (2).
$3 x - 2 (- 0.5) = 5.5$
Solve for $x$ .
$3x+1=5.53x=4.5x=1.5\begin{aligned} 3x+1&=5.5\\ 3x&=4.5\\ x&=1.5 \end{aligned}$
Give the answer.
$x=1.5,y=−0.5x=1.5,\quad y=-0.5$

Word problems

In word problems, you choose variables first. Then translate the information into equations.

Definition

Variable

A variable is a letter that represents an unknown value. For example, $c$ could represent the price of one coffee in pence.

Example

Prices in a shop

A café sells drinks and biscuits. Two drinks and five biscuits cost £8.90. One drink and three biscuits cost £4.95. Find the price of one drink and one biscuit.

Let $d$ be the price of one drink in pence, and let $b$ be the price of one biscuit in pence.
Write an equation for each sentence.
$2d+5b=890(1)d+3b=495(2)\begin{aligned} 2d+5b&=890 \quad \text{(1)}\\ d+3b&=495 \quad \text{(2)} \end{aligned}$
Multiply equation (2) by 2 so the $d$ coefficients match.
$\quad \text{(3)}$
Subtract equation (1) from equation (3).
$(2d+6b)−(2d+5b)=990−890b=100\begin{aligned} (2d+6b)-(2d+5b)&=990-890\\ b&=100 \end{aligned}$
Substitute $b = 100$ into equation (2).
$d + 3 (100) = 495$
Solve for $d$ .
$d+300=495d=195\begin{aligned} d+300&=495\\ d&=195 \end{aligned}$
Convert from pence to pounds.
$d=195,b=100d=195,\quad b=100$
The drink costs £1.95 and the biscuit costs £1.00.

Common Mistake

Remember the units

If you use pence in your equations, your final numbers are in pence. Convert back to pounds if the question asks for prices.

Exam technique

In the exam

Line up the equations neatly and decide which variable you want to eliminate.
If the coefficients do not match, multiply a whole equation first.
After finding one variable, substitute it back into one original equation and give both values clearly.

Self review

Check yourself

Can you explain when to add equations and when to subtract them?
If $3 x = 7 + y$ , can you rearrange it into standard form?
In a word problem, what should you define before writing the equations?

Simultaneous Equations

What you'll learn

What simultaneous equations are and what a solution means.
How to solve linear simultaneous equations by elimination.
How to handle rearranged equations and decimals.
How to turn short word problems into equations.

The basic idea

A linear equation is an equation where the variables only have power 1, such as $4 x + 3 y = 18$ . Its graph is a straight line.

Definition

Simultaneous equations

Simultaneous equations are two or more equations that are true at the same time. For IGCSE Grade 5, you usually solve two linear equations with two variables, often $x$ and $y$ .

A solution is the pair of values that makes both equations true. For example, $x = 2$ and $y = 3$ is written as the solution pair $(2, 3)$ .

Graphically, each linear equation is a straight line. The simultaneous solution is where the two lines cross.

Two straight-line graphs crossing at the labelled solution point (2, 3)

Key Idea

The big picture

Solving simultaneous equations means finding the one value of $x$ and the one value of $y$ that work in both equations.

Example

Checking whether a pair is a solution

Check whether $x = 4$ and $y = 1$ solves the equations $2 x + 3 y = 11$ and $5 x - y = 19$ .

Substitute $x = 4$ and $y = 1$ into the first equation.
$2 (4) + 3 (1) = 8 + 3 = 11$
The first equation works because the left side equals 11.
Substitute $x = 4$ and $y = 1$ into the second equation.
$5 (4) - 1 = 20 - 1 = 19$
The second equation also works, so $x = 4$ and $y = 1$ is the solution.

The elimination method

The main Grade 5 method is elimination.

Definition

Eliminate

To eliminate a variable means to remove it from the equations by adding or subtracting the equations.

You try to make either the $x$ terms or the $y$ terms cancel out. Then you solve for the remaining variable.

When the signs are opposite: add

If one equation has $+ 3 y$ and the other has $- 3 y$ , adding the equations cancels the $y$ terms.

Example

Eliminating by adding

Solve the simultaneous equations $4 x + 3 y = 26$ and $x - 3 y = - 1$ .

Write the equations clearly and label them.
$4x+3y=26(1)x−3y=−1(2)\begin{aligned} 4x+3y&=26 \quad \text{(1)}\\ x-3y&=-1 \quad \text{(2)} \end{aligned}$
The $y$ terms are opposites, $+ 3 y$ and $- 3 y$ , so add the two equations.
$4x+x=5x3y−3y=026+(−1)=25\begin{aligned} 4x+x&=5x\\ 3y-3y&=0\\ 26+(-1)&=25 \end{aligned}$
Solve the new equation.
$5 x = 25$
Divide by 5.
$x = 5$
Substitute $x = 5$ into equation (2).
$5 - 3 y = - 1$
Solve for $y$ .
$−3y=−6y=2\begin{aligned} -3y&=-6\\ y&=2 \end{aligned}$
Give both values.
$x=5,y=2x=5,\quad y=2$

Tip

Add or subtract?

If the matching terms have opposite signs, add the equations. If they have the same sign, subtract the equations.

When the signs are the same: subtract

If both equations have the same $x$ coefficient, such as $2 x$ and $2 x$ , subtracting can eliminate $x$ .

Example

Eliminating by subtracting

Solve the simultaneous equations $2 x + 5 y = 28$ and $2 x - y = 4$ .

Label the equations.
$2x+5y=28(1)2x−y=4(2)\begin{aligned} 2x+5y&=28 \quad \text{(1)}\\ 2x-y&=4 \quad \text{(2)} \end{aligned}$
The $x$ terms are the same, so subtract equation (2) from equation (1).
$(2x+5y)−(2x−y)=28−46y=24\begin{aligned} (2x+5y)-(2x-y)&=28-4\\ 6y&=24 \end{aligned}$
Solve for $y$ .
$y = 4$
Substitute $y = 4$ into equation (2).
$2 x - 4 = 4$
Solve for $x$ .
$2x=8x=4\begin{aligned} 2x&=8\\ x&=4 \end{aligned}$
State the solution.
$x=4,y=4x=4,\quad y=4$

Common Mistake

Subtracting negatives incorrectly

Be careful with signs. In the example above, subtracting $- y$ means $5 y - (- y) = 6 y$ , not $4 y$ .

Making a coefficient match

Sometimes neither variable cancels straight away. A coefficient is the number multiplying a variable, such as the 4 in $4 x$ .

To eliminate a variable, you can multiply one whole equation so that a coefficient matches.

Example

Multiplying first, then eliminating

Solve $2 x - 3 y = 4$ and $4 x - y = 18$ .

Label the equations.
$2x−3y=4(1)4x−y=18(2)\begin{aligned} 2x-3y&=4 \quad \text{(1)}\\ 4x-y&=18 \quad \text{(2)} \end{aligned}$
The first equation has $2 x$ and the second has $4 x$ . Multiply equation (1) by 2 so the $x$ coefficients match.
$\quad \text{(3)}$
Now subtract equation (2) from equation (3).
$(4x−6y)−(4x−y)=8−18−5y=−10\begin{aligned} (4x-6y)-(4x-y)&=8-18\\ -5y&=-10 \end{aligned}$
Solve for $y$ .
$y = 2$
Substitute $y = 2$ into equation (2).
$4 x - 2 = 18$
Solve for $x$ .
$4x=20x=5\begin{aligned} 4x&=20\\ x&=5 \end{aligned}$
Write the final answer.
$x=5,y=2x=5,\quad y=2$

Common Mistake

Only multiplying part of an equation

If you multiply an equation by 2, multiply every term on both sides. For example, $2 x - 3 y = 4$ becomes $4 x - 6 y = 8$ .

Rearranged equations

Not every question starts in the neat form $a x + b y = c$ . You may first need to rearrange.

Definition

Standard form

For this topic, standard form means writing a linear equation like $a x + b y = c$ , with the $x$ and $y$ terms on the left and the number on the right.

Example

Rearranging before solving

Solve $3 x = 10 + y$ and $x + 4 y = 12$ .

Rearrange the first equation by subtracting $y$ from both sides.
$3 x - y = 10$
Label the two equations.
$3x−y=10(1)x+4y=12(2)\begin{aligned} 3x-y&=10 \quad \text{(1)}\\ x+4y&=12 \quad \text{(2)} \end{aligned}$
Multiply equation (2) by 3 so the $x$ coefficients match.
$\quad \text{(3)}$
Subtract equation (1) from equation (3).
$(3x+12y)−(3x−y)=36−1013y=26\begin{aligned} (3x+12y)-(3x-y)&=36-10\\ 13y&=26 \end{aligned}$
Solve for $y$ .
$y = 2$
Substitute $y = 2$ into equation (2).
$x + 8 = 12$
Solve for $x$ .
$x = 4$
State the solution.
$x=4,y=2x=4,\quad y=2$

Decimals in simultaneous equations

Decimals do not change the method. You can either work with decimals carefully or multiply everything by 10 to clear one decimal place.

Tip

Clearing decimals

If an equation contains 4.5, multiplying the whole equation by 10 changes it to 45. This can make elimination easier.

Example

Solving with decimals

Solve $6 x + 5 y = 6.5$ and $3 x - 2 y = 5.5$ .

Label the equations.
$6x+5y=6.5(1)3x−2y=5.5(2)\begin{aligned} 6x+5y&=6.5 \quad \text{(1)}\\ 3x-2y&=5.5 \quad \text{(2)} \end{aligned}$
Multiply equation (2) by 2 so the $x$ coefficients match.
$\quad \text{(3)}$
Subtract equation (3) from equation (1).
$(6x+5y)−(6x−4y)=6.5−119y=−4.5\begin{aligned} (6x+5y)-(6x-4y)&=6.5-11\\ 9y&=-4.5 \end{aligned}$
Solve for $y$ .
$y = - 0.5$
Substitute $y = - 0.5$ into equation (2).
$3 x - 2 (- 0.5) = 5.5$
Solve for $x$ .
$3x+1=5.53x=4.5x=1.5\begin{aligned} 3x+1&=5.5\\ 3x&=4.5\\ x&=1.5 \end{aligned}$
Give the answer.
$x=1.5,y=−0.5x=1.5,\quad y=-0.5$

Word problems

In word problems, you choose variables first. Then translate the information into equations.

Definition

Variable

A variable is a letter that represents an unknown value. For example, $c$ could represent the price of one coffee in pence.

Example

Prices in a shop

A café sells drinks and biscuits. Two drinks and five biscuits cost £8.90. One drink and three biscuits cost £4.95. Find the price of one drink and one biscuit.

Let $d$ be the price of one drink in pence, and let $b$ be the price of one biscuit in pence.
Write an equation for each sentence.
$2d+5b=890(1)d+3b=495(2)\begin{aligned} 2d+5b&=890 \quad \text{(1)}\\ d+3b&=495 \quad \text{(2)} \end{aligned}$
Multiply equation (2) by 2 so the $d$ coefficients match.
$\quad \text{(3)}$
Subtract equation (1) from equation (3).
$(2d+6b)−(2d+5b)=990−890b=100\begin{aligned} (2d+6b)-(2d+5b)&=990-890\\ b&=100 \end{aligned}$
Substitute $b = 100$ into equation (2).
$d + 3 (100) = 495$
Solve for $d$ .
$d+300=495d=195\begin{aligned} d+300&=495\\ d&=195 \end{aligned}$
Convert from pence to pounds.
$d=195,b=100d=195,\quad b=100$
The drink costs £1.95 and the biscuit costs £1.00.

Common Mistake

Remember the units

If you use pence in your equations, your final numbers are in pence. Convert back to pounds if the question asks for prices.

Exam technique

In the exam

Line up the equations neatly and decide which variable you want to eliminate.
If the coefficients do not match, multiply a whole equation first.
After finding one variable, substitute it back into one original equation and give both values clearly.

Self review

Check yourself

Can you explain when to add equations and when to subtract them?
If $3 x = 7 + y$ , can you rearrange it into standard form?
In a word problem, what should you define before writing the equations?

Simultaneous Equations

What you'll learn

The basic idea

The elimination method

When the signs are opposite: add

When the signs are the same: subtract

Making a coefficient match

Rearranged equations

Decimals in simultaneous equations

Word problems

Recap questions

You've reached the end

Simultaneous Equations

What you'll learn

The basic idea

The elimination method

When the signs are opposite: add

When the signs are the same: subtract

Making a coefficient match

Rearranged equations

Decimals in simultaneous equations

Word problems

Recap questions

You've reached the end

Simultaneous Equations

What you'll learn

The basic idea

The elimination method

When the signs are opposite: add

When the signs are the same: subtract

Making a coefficient match

Rearranged equations

Decimals in simultaneous equations

Word problems

Recap questions

You've reached the end

Simultaneous Equations Revision Guide

Simultaneous Equations

What you'll learn

The basic idea

The elimination method

When the signs are opposite: add

When the signs are the same: subtract

Making a coefficient match

Rearranged equations

Decimals in simultaneous equations

Word problems

Recap questions

You've reached the end

Simultaneous Equations Revision Guide