Year 9 · Chapter 5

Simultaneous equations and inequalities

By the end of this chapter you will be able to

Solve simultaneous equations by elimination, multiplying one or both equations first
Solve simultaneous equations by substitution when one equation gives y in terms of x
Read the solution of two simultaneous equations from the point where two graphs cross, and check it
Draw and shade the region for one or more linear inequalities, using dashed or solid boundaries
Count the integer points that satisfy several inequalities, and form and solve equations from a story

You should already know

Solving one-step and two-step linear equations (Year 8, Chapter 6)
Drawing straight-line graphs and finding the gradient and intercept (Year 8, Chapter 7)
Solving inequalities with one variable and showing them on a number line (Year 8, Chapter 6)

Where this leads

Quadratic equations and their graphs (Year 9, Chapters 4 and 6)
Simultaneous equations with a quadratic and linear equation (GCSE)
Linear programming and regions in GCSE algebra

National Curriculum for England, Key Stage 3 mathematics, Year 9. Content: Solve two linear simultaneous equations algebraically by elimination and by substitution, including fractional answers, Solve simultaneous equations graphically, and interpret the point of intersection, Solve linear inequalities in two variables, and show the solution set by shading a region on a graph, Find the integer points that satisfy several inequalities at once, and form simultaneous equations from a context.

Exercise P Do you remember?

Non-calculator

FSolve 2x + 5 = 13.

Done

FWhat is the y-intercept of the line y = 3x + 5?

Done

FWhat is the gradient of the line y = −2x + 1?

Done

FIs (3, 2) a solution of x + y = 5?

Done

FWhich point is on the line y = x + 1 when x = 2?

Done

5.1

Solving simultaneous equations by elimination

Year 9
Solve two linear simultaneous equations by elimination

Video lesson: Solving simultaneous equations by eliminationComing soon

At a cinema, 2 adult tickets and 3 child tickets cost one total, and 1 adult and 3 children cost another. Two equations with two unknowns describe the situation. Elimination removes one unknown so that the other can be found.

Simultaneous equations are two equations that are both true at the same time. Their solution is the pair of values of x and y that works in both equations.

In elimination, multiply one or both equations so that one unknown has the same coefficient in both. Then add or subtract the equations to remove that unknown. Solve the remaining equation, then substitute to find the other unknown.

Key ideas

Label the equations (1) and (2).
Make the coefficients of one unknown match, multiplying where necessary.
Subtract (or add) to eliminate that unknown, then solve for the other.
Substitute back into one of the original equations, and check in the other.

Worked example 5.1.1

Solve 3x + 2y = 16 and x + 4y = 22.

Method. Multiply the equations so the x terms match, then subtract to eliminate x.

Write the equations and label them. Give each equation a label so you can refer to it.
(1) 3x + 2y = 16
(2) x + 4y = 22
Make the x coefficients the same. Multiply (1) by 1 and (2) by 3, so both x terms become 3x.
(1) × 1: 3x + 2y = 16
(2) × 3: 3x + 12y = 66
Subtract to eliminate x. Subtracting removes the x terms, leaving an equation in y only.
(3x + 2y = 16) − (3x + 12y = 66)
-10y = -50
Solve for y. Divide both sides by the number in front of y.
y = -50 ÷ -10 = 5
Substitute to find x. Put y into equation (1) and solve for x.
3x + 2 × (5) = 16, so x = 2
Write the answer.
x = 2, y = 5

Check. Check in equation (2): 1(2) + 4(5) = 22, which should equal 22.

Watch out. Subtract the equations carefully, one term at a time. Watch the minus signs.

Worked example 5.1.2

Solve 3x + 2y = 4 and x + 4y = 1. Give your answers as fractions where needed.

Method. Multiply the equations so the x terms match, then subtract to eliminate x.

Write the equations and label them. Give each equation a label so you can refer to it.
(1) 3x + 2y = 4
(2) x + 4y = 1
Make the x coefficients the same. Multiply (1) by 1 and (2) by 3, so both x terms become 3x.
(1) × 1: 3x + 2y = 4
(2) × 3: 3x + 12y = 3
Subtract to eliminate x. Subtracting removes the x terms, leaving an equation in y only.
(3x + 2y = 4) − (3x + 12y = 3)
-10y = 1
Solve for y. Divide both sides by the number in front of y.
y = 1 ÷ -10 = −1/10
Substitute to find x. Put y into equation (1) and solve for x.
3x + 2 × (−1/10) = 4, so x = 7/5
Write the answer.
x = 7/5, y = −1/10

Check. Check in equation (2): 1(7/5) + 4(−1/10) = 1, which should equal 1.

Watch out. Subtract the equations carefully, one term at a time. Watch the minus signs.

Worked example 5.1.3

Solve 2x + 3y = 12 and 2x − y = 2.

Method. Multiply the equations so the x terms match, then subtract to eliminate x.

Write the equations and label them. Give each equation a label so you can refer to it.
(1) 2x + 3y = 12
(2) 2x − y = 2
Make the x coefficients the same. Multiply (1) by 1 and (2) by 1, so both x terms become 2x.
(1) × 1: 2x + 3y = 12
(2) × 1: 2x − y = 2
Subtract to eliminate x. Subtracting removes the x terms, leaving an equation in y only.
(2x + 3y = 12) − (2x − y = 2)
4y = 10
Solve for y. Divide both sides by the number in front of y.
y = 10 ÷ 4 = 5/2
Substitute to find x. Put y into equation (1) and solve for x.
2x + 3 × (5/2) = 12, so x = 9/4
Write the answer.
x = 9/4, y = 5/2

Check. Check in equation (2): 2(9/4) + -1(5/2) = 2, which should equal 2.

Watch out. Subtract the equations carefully, one term at a time. Watch the minus signs.

Watch out

Watch out: when you subtract an equation, subtract every term, including the right-hand side. A common error is to forget the sign of a term that was negative.

Exam tip

Exam tip. Check your answer in the equation you did not use. If it does not work, there is a mistake earlier.

5.2

Solving simultaneous equations by substitution

Year 9
Solve two linear simultaneous equations by substitution

Video lesson: Solving simultaneous equations by substitutionComing soon

A shop charges a fixed price per pencil. The first equation gives the price of one pencil in terms of the number of pencils. The second equation is another total. Substitution uses the first equation to replace one unknown in the second.

When one equation is already in the form y = (something), substitute that expression for y in the other equation. The new equation has only x, so solve it for x. Then find y from the first equation.

Always put the whole expression in brackets before you multiply it.

Key ideas

Substitute the expression for y into the other equation.
Use brackets: 2(x + 3) is not 2x + 3.
Solve for x, then substitute back to find y.
Check the answer in both original equations.

Worked example 5.2.1

Solve y = 2x + 1 and 3x + y = 11.

Method. Substitute the expression for y into the other equation, then solve for x.

Write the two equations. The first equation already gives y in terms of x.
(1) y = 2x + 1
(2) 3x + y = 11
Substitute y into equation (2). Replace every y in equation (2) by the expression from (1).
3x + 1(2x + 1) = 11
Expand and collect the x terms. Multiply the bracket out, then gather the x terms on one side.
5x + 1 = 11
Solve for x. Move the number to the right, then divide.
5x = 10, so x = 2
Find y. Put x back into equation (1).
y = 2 × (2) + 1 = 5
Write the answer.
x = 2, y = 5

Check. Check in equation (2): 3(2) + 1(5) = 11, which should equal 11.

Watch out. Put the whole expression in brackets before you multiply. Forgetting the bracket is a common error.

Worked example 5.2.2

Solve y = 3x − 2 and x + 2y = 10.

Method. Substitute the expression for y into the other equation, then solve for x.

Write the two equations. The first equation already gives y in terms of x.
(1) y = 3x − 2
(2) x + 2y = 10
Substitute y into equation (2). Replace every y in equation (2) by the expression from (1).
1x + 2(3x − 2) = 10
Expand and collect the x terms. Multiply the bracket out, then gather the x terms on one side.
7x − 4 = 10
Solve for x. Move the number to the right, then divide.
7x = 14, so x = 2
Find y. Put x back into equation (1).
y = 3 × (2) − 2 = 4
Write the answer.
x = 2, y = 4

Check. Check in equation (2): 1(2) + 2(4) = 10, which should equal 10.

Watch out. Put the whole expression in brackets before you multiply. Forgetting the bracket is a common error.

Worked example 5.2.3

Solve y = 2x and 3x + y = 12. Give your answers as fractions.

Method. Substitute the expression for y into the other equation, then solve for x.

Write the two equations. The first equation already gives y in terms of x.
(1) y = 2x
(2) 3x + y = 12
Substitute y into equation (2). Replace every y in equation (2) by the expression from (1).
3x + 1(2x) = 12
Expand and collect the x terms. Multiply the bracket out, then gather the x terms on one side.
5x = 12
Solve for x. Move the number to the right, then divide.
5x = 12, so x = 12/5
Find y. Put x back into equation (1).
y = 2 × (12/5) = 24/5
Write the answer.
x = 12/5, y = 24/5

Check. Check in equation (2): 3(12/5) + 1(24/5) = 12, which should equal 12.

Watch out. Put the whole expression in brackets before you multiply. Forgetting the bracket is a common error.

Watch out

Watch out: in 3x + 2(2x + 1) the 2 multiplies both terms in the bracket. It gives 3x + 4x + 2, not 3x + 4x + 1.

Exam tip

Exam tip. Substitution is quicker than elimination when one equation is already in the form y = ....

5.3

Solving simultaneous equations graphically

Year 9
Solve simultaneous equations graphically, and interpret the point of intersection

Video lesson: Solving simultaneous equations graphicallyComing soon

Two mobile phone plans are drawn as straight lines on a graph of cost against minutes. The point where the lines cross is the number of minutes at which both plans cost the same. That point is the solution of both equations.

The solution of two simultaneous equations is the point of intersection of their graphs. The x-coordinate gives the value of x, and the y-coordinate gives the value of y. For exact answers, use algebra and check on the graph.

Key ideas

The point of intersection is the solution of both equations.
Read the coordinates from the graph, then check them in both equations.
Parallel lines do not cross, so there is no solution.
Use the graph to check your algebra, not as the only method.

Worked example 5.3.1

The lines y = x + 1 and y = −2x + 7 are drawn. Find where they cross.

Method. Find the point of intersection of the two lines. Then check it in both equations.

Find the point where the lines cross. The point of intersection lies on both lines, so it is the solution of both equations.
y = 1x +1 and y = −2x +7
Read the coordinates. The x-coordinate is the x value and the y-coordinate is the y value at that point.
the lines cross at (2, 3)
Check in both equations. Substitute the point into each equation to make sure it works.
3 = 1 × (2) +1 and 3 = −2 × (2) +7
Write the answer.
x = 2, y = 3

Check. Both equations give y = 3 when x = 2.

Watch out. Read the point from the graph only to check your algebra. Exact answers come from the equations.

Worked example 5.3.2

Find the point of intersection of y = 2x − 1 and y = −x + 5.

Method. Find the point of intersection of the two lines. Then check it in both equations.

Find the point where the lines cross. The point of intersection lies on both lines, so it is the solution of both equations.
y = 2x −1 and y = −1x +5
Read the coordinates. The x-coordinate is the x value and the y-coordinate is the y value at that point.
the lines cross at (2, 3)
Check in both equations. Substitute the point into each equation to make sure it works.
3 = 2 × (2) −1 and 3 = −1 × (2) +5
Write the answer.
x = 2, y = 3

Check. Both equations give y = 3 when x = 2.

Watch out. Read the point from the graph only to check your algebra. Exact answers come from the equations.

Worked example 5.3.3

Find the point of intersection of y = 3x − 5 and y = x − 1.

Method. Find the point of intersection of the two lines. Then check it in both equations.

Find the point where the lines cross. The point of intersection lies on both lines, so it is the solution of both equations.
y = 3x −5 and y = 1x −1
Read the coordinates. The x-coordinate is the x value and the y-coordinate is the y value at that point.
the lines cross at (2, 1)
Check in both equations. Substitute the point into each equation to make sure it works.
1 = 3 × (2) −5 and 1 = 1 × (2) −1
Write the answer.
x = 2, y = 1

Check. Both equations give y = 1 when x = 2.

Watch out. Read the point from the graph only to check your algebra. Exact answers come from the equations.

Watch out

Watch out: a graph read by eye is only approximate. Check exact values in both equations.

Exam tip

Exam tip. Label each line with its equation, and mark the point clearly on the graph.

5.4

Inequalities in two variables

Year 9
Solve linear inequalities in two variables, and show the solution set by shading a region

Video lesson: Inequalities in two variablesComing soon

A school canteen wants to sell at least 2 burgers and at most 6 meals in total. Each condition is a linear inequality. The values that satisfy both conditions form a region on a graph.

A linear inequality such as y > x + 1 has a boundary line: y = x + 1. Use a dashed line for < or >, and a solid line for ≤ or ≥, because a solid line includes the boundary.

To decide which side to shade, test a point such as (0, 0). If it satisfies the inequality, shade the side that contains it.

Key ideas

Replace the inequality sign with = to draw the boundary line.
Dashed line for < and >; solid line for ≤ and ≥.
Test a point not on the line, such as (0, 0), to choose the side to shade.
Several inequalities: shade only the region that satisfies all of them.

Worked example 5.4.1

Show the region for y ≥ x + 1 by shading.

Method. Draw the boundary lines, test a point on each side, and shade the region that is true for all the inequalities.

Draw each boundary line. Replace the inequality sign with = to get the boundary. Use a dashed line for < or >, and a solid line for ≤ or ≥.
y = x + 1
Test a point. Choose a point that is not on a line, such as (0, 0), and check whether it satisfies the inequality.
If the test point works, shade the side that contains it.
Shade the region. The region that satisfies all the inequalities is the one that is shaded on every test.
Shade the region with corners near (-1, 0), (-1, 6), (5, 6)
Write the answer.
Shade the region that satisfies all of the inequalities

Check. A point inside the shaded region should satisfy every inequality.

Watch out. Use a dashed line when the inequality is strict (< or >). A solid line includes the boundary.

Worked example 5.4.2

Show the region for x + y < 4 on a graph.

Method. Draw the boundary lines, test a point on each side, and shade the region that is true for all the inequalities.

Draw each boundary line. Replace the inequality sign with = to get the boundary. Use a dashed line for < or >, and a solid line for ≤ or ≥.
x + y = 4
Test a point. Choose a point that is not on a line, such as (0, 0), and check whether it satisfies the inequality.
If the test point works, shade the side that contains it.
Shade the region. The region that satisfies all the inequalities is the one that is shaded on every test.
Shade the region with corners near (-1, -1), (-1, 5), (5, -1)
Write the answer.
Shade the region that satisfies all of the inequalities

Check. A point inside the shaded region should satisfy every inequality.

Watch out. Use a dashed line when the inequality is strict (< or >). A solid line includes the boundary.

Worked example 5.4.3

Show the region for x ≥ 1 and y < 3 on a graph.

Method. Draw the boundary lines, test a point on each side, and shade the region that is true for all the inequalities.

Draw each boundary line. Replace the inequality sign with = to get the boundary. Use a dashed line for < or >, and a solid line for ≤ or ≥.
x = 1
y = 3
Test a point. Choose a point that is not on a line, such as (0, 0), and check whether it satisfies the inequality.
If the test point works, shade the side that contains it.
Shade the region. The region that satisfies all the inequalities is the one that is shaded on every test.
Shade the region with corners near (1, -1), (1, 3), (6, -1)
Write the answer.
Shade the region that satisfies all of the inequalities

Check. A point inside the shaded region should satisfy every inequality.

Watch out. Use a dashed line when the inequality is strict (< or >). A solid line includes the boundary.

Watch out

Watch out: a point on the boundary of a strict inequality (< or >) is not part of the solution. Use a dashed line.

Exam tip

Exam tip. Check your shading with one test point inside the shaded region.

5.5

Integer points in a region

Year 9
Find the integer points that satisfy several inequalities at once

Video lesson: Integer points in a regionComing soon

A party planner invites x adults and y children. The numbers must be whole numbers, and the rules for the party give several inequalities. Only the whole-number points that satisfy every rule are possible invitations.

An integer point has whole-number coordinates. To find the integer points that satisfy several inequalities, list the points that satisfy each one, or test points systematically in a table, and keep only those that satisfy all of them.

Key ideas

Integer points have whole-number x and y.
A point must satisfy every inequality, not just one.
Test the points in a systematic table, row by row.
Strict inequalities (< or >) exclude the boundary.

Worked example 5.5.1

Find the number of integer points that satisfy x ≥ 0, y ≥ 0 and x + y ≤ 3.

Method. Test each x value in turn and count the y values that fit.

Start with x = 0. Then y must satisfy y ≥ 0 and y ≤ 3.
y = 0, 1, 2, 3: four points
Try x = 1. Then y ≤ 2, so y = 0, 1, 2.
three points
Try x = 2. Then y ≤ 1, so y = 0, 1.
two points
Try x = 3. Then y = 0, so one point.
one point
Add the counts.
4 + 3 + 2 + 1 = 10

Check. There are 10 integer points, for example (0, 0), (1, 2) and (3, 0).

Worked example 5.5.2

How many integer points satisfy x ≥ 1, y ≥ 1 and x + y ≤ 4?

Method. Test each x value in turn and count the y values that fit.

Start with x = 1. Then y ≥ 1 and y ≤ 3, so y = 1, 2, 3.
3 points
Try x = 2. Then y ≥ 1 and y ≤ 2, so y = 1, 2.
2 points
Try x = 3. Then y = 1 only.
1 point
Add the counts.
3 + 2 + 1 = 6

Check. The points are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2) and (3, 1).

Worked example 5.5.3

How many integer points satisfy x ≥ 0, y ≥ 0 and x + y ≤ 2?

Method. Count the points for each x value.

For x = 0. Then y = 0, 1, 2: three points.
3
For x = 1. Then y = 0, 1: two points.
2
For x = 2. Then y = 0: one point.
1
Add the counts.
3 + 2 + 1 = 6

Check. The six points are (0, 0), (0, 1), (0, 2), (1, 0), (1, 1) and (2, 0).

Watch out

Watch out: check the boundary carefully. The points (1, 2) and (2, 1) are on x + y = 3, so they count when the inequality is ≤.

Exam tip

Exam tip. Draw a quick table of x values and count the y values for each, to avoid missing points.

Exercise 5E Integer points in a region

Non-calculator

PSHow many integer points satisfy x ≥ 0, y ≥ 0 and x + y ≤ 4?

Done

PSHow many integer points satisfy x ≥ 1, y ≥ 1 and x + y ≤ 5?

Done

PSHow many integer points satisfy y ≤ x, y ≥ 0 and x ≤ 3?

Done

PSHow many integer points satisfy y < 2x, y ≥ 0 and x ≤ 3?

Done

PSHow many integer points satisfy 2x + y ≤ 6, x ≥ 0 and y ≥ 0?

Done

PSHow many integer points satisfy x + y ≥ 3, x ≤ 2 and y ≤ 2?

Done

PSHow many integer points satisfy x + 2y ≤ 6, x ≥ 0 and y ≥ 0?

Done

PSHow many integer points satisfy x + y < 5, x ≥ 0 and y ≥ 0?

Done

5.6

Forming simultaneous equations from a context

Year 9
Form simultaneous equations from a context and solve them

Video lesson: Forming simultaneous equations from a contextComing soon

Sam buys tickets for a school trip. Two different totals are given for different numbers of tickets. Each total gives an equation, and solving the two equations gives the price of each ticket.

Choose letters for the two unknown quantities, write one equation for each piece of information, then solve them. Finally, answer the question in words with the correct unit.

Key ideas

Let each unknown be a letter, and say what it stands for.
Each piece of information gives one equation.
Solve by elimination or substitution.
Answer the question asked, with the unit, and check it makes sense.

Worked example 5.6.1

Adult and child tickets: 3 adults and 2 children cost £27, and 1 adult and 4 children cost £19. Find the adult ticket price.

Method. Form two equations, eliminate one unknown, then answer the question in context.

Write the two equations. Let x be the adult price and y be the child price. Use the information to write two equations.
(1) 3x + 2y = 27
(2) x + 4y = 19
Make the x coefficients the same. Multiply (1) by 1 and (2) by 3.
3x + 2y = 27
3x + 12y = 57
Subtract to eliminate x. The x terms cancel, leaving an equation in y.
-10y = -30, so y = 3
Find x. Substitute y into equation (1).
x = 7
Answer the question in context. The question asks for the adult price.
£7

Check. Check both equations with x = 7 and y = 3.

Watch out. Give the answer with the correct unit, and check that it makes sense in the story.

Worked example 5.6.2

4 pens and 1 ruler cost £6, and 2 pens and 3 rulers cost £8. Find the cost of one pen.

Method. Form two equations, eliminate one unknown, then answer the question in context.

Write the two equations. Let x be the pen cost and y be the ruler cost. Use the information to write two equations.
(1) 4x + y = 6
(2) 2x + 3y = 8
Make the x coefficients the same. Multiply (1) by 1 and (2) by 2.
4x + 1y = 6
4x + 6y = 16
Subtract to eliminate x. The x terms cancel, leaving an equation in y.
-5y = -10, so y = 2
Find x. Substitute y into equation (1).
x = 1
Answer the question in context. The question asks for the pen cost.
£1

Check. Check both equations with x = 1 and y = 2.

Watch out. Give the answer with the correct unit, and check that it makes sense in the story.

Worked example 5.6.3

A bag of apples and pears: 2 apples and 3 pears cost £13, and 1 apple and 1 pear cost £5. Find the cost of one pear.

Method. Form two equations, eliminate one unknown, then answer the question in context.

Write the two equations. Let x be the apple cost and y be the pear cost. Use the information to write two equations.
(1) 2x + 3y = 13
(2) x + y = 5
Make the x coefficients the same. Multiply (1) by 1 and (2) by 2.
2x + 3y = 13
2x + 2y = 10
Subtract to eliminate x. The x terms cancel, leaving an equation in y.
1y = 3, so y = 3
Find x. Substitute y into equation (1).
x = 2
Answer the question in context. The question asks for the pear cost.
£3

Check. Check both equations with x = 2 and y = 3.

Watch out. Give the answer with the correct unit, and check that it makes sense in the story.

Watch out

Watch out: write the equations from the story before you solve them. Check that each equation matches the total it describes.

Exam tip

Exam tip. Write what x and y stand for at the start, with units, so the answer has the right unit.

Chapter summary

Skill	Method
Elimination	Make the coefficients of one unknown match, then add or subtract the equations.
Substitution	Put the expression for y into the other equation, solve for x, then find y.
Graphs	The point where two lines cross is the solution. Check it in both equations.
Inequalities	Boundary line from =. Dashed for < or >, solid for ≤ or ≥. Test a point to choose the side.
Integer points	Test each whole-number point against every inequality, in a systematic table.
Context	Let x and y stand for the quantities, write two equations, solve, and answer with units.

Can you…?

I can solve two simultaneous equations by elimination
I can solve two simultaneous equations by substitution
I can read the solution of two equations from where their graphs cross
I can draw and shade the region for linear inequalities, using dashed or solid lines
I can count the integer points that satisfy several inequalities
I can form two equations from a story and solve them

Key terms

simultaneous equations: two equations solved together, with the same values of x and y in both
elimination: removing one unknown by adding or subtracting equations
point of intersection: the point where two graphs cross
boundary line: the line you get by replacing the inequality sign with =
integer point: a point where both coordinates are whole numbers

Total 26 marks. Allow about 26 minutes.

← Chapter 4: Quadratic expressions and equations Chapter 6: Functions and graphs: quadratic, cubic and reciprocal →

Simultaneous equations and inequalities

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Solving simultaneous equations by elimination

Key ideas

Watch out

Exam tip

Exercise 5A Solving by elimination

Solving simultaneous equations by substitution

Key ideas

Watch out

Exam tip

Exercise 5B Solving by substitution

Solving simultaneous equations graphically

Key ideas

Watch out

Exam tip

Exercise 5C Solving graphically

Inequalities in two variables

Key ideas

Watch out

Exam tip

Exercise 5D Shading regions

Integer points in a region

Key ideas

Watch out

Exam tip

Exercise 5E Integer points in a region

Forming simultaneous equations from a context

Key ideas

Watch out

Exam tip

Exercise 5F Forming and solving simultaneous equations

Chapter summary

Can you…?

Key terms

Exercise 5Z Check-up

Exercise 5X Practice test questions