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KS3 Maths · Year 8Chapter 6 · Linear equations, inequalities and simultaneous basics
Year 8 · Chapter 6

Linear equations, inequalities and simultaneous basics

By the end of this chapter you will be able to

  • Solve one-step and two-step equations, and equations with the unknown on both sides
  • Solve equations with brackets and with fractions, by clearing the fraction first
  • Write an equation from a short story, and solve it to answer the question
  • Solve an inequality and show the answer on a number line, with an open or closed circle
  • Solve two simultaneous equations with integer solutions by elimination or substitution, and check both equations

You should already know

  • Solving one-step equations such as x + 4 = 9 and 3x = 12 (Year 7)
  • Simplifying expressions and expanding a single bracket (Year 8, Chapter 5)
  • Coordinates, and reading a number line (Year 7)

Where this leads

  • Solving equations with fractions and quadratics, and rearranging formulae (Year 9)
  • Graphical solutions of simultaneous equations and quadratic equations (Year 9 and GCSE)
  • Inequalities in two variables and regions on a graph (GCSE)

National Curriculum for England, Key Stage 3 mathematics, Year 8. Content: Solve linear equations in one unknown, including brackets, unknowns on both sides and fractions, Form and solve an equation from a context, Solve linear inequalities in one variable, show them on a number line, and list integer solutions, Solve two simultaneous linear equations by elimination or substitution.

Exercise P  Do you remember?

Non-calculator
1
FSolve x + 4 = 9.
2
FSolve 3x = 12.
3
FExpand 2(x + 5).
4
FWhich number is bigger, 3 or −2?
5
FWork out the value of 4x − 3 when x = 5.
6.1

Solving one-step and two-step equations

Year 8
Solve linear equations in one unknown, Solve equations with the unknown on both sides
Video lesson: Solving one-step and two-step equationsComing soon
Ali has some sweets in a bag. He has 3 bags of the same number, plus 5 loose sweets, and he has 20 sweets altogether. He writes the equation 3x + 5 = 20 to find how many sweets are in one bag.

An equation is like a balance. Whatever you do to one side, you must do to the other side, so that it stays balanced.

To solve a two-step equation, undo the number terms first, then divide by the number in front of x.

If there are unknowns on both sides, first collect the x terms on one side, so that only numbers are on the other side.

Key ideas

  • Do the same to both sides of the equation.
  • Undo addition and subtraction first, then undo multiplication and division.
  • Collect the x terms on one side before you solve.
  • Check your answer by substituting it into the original equation.
Worked example 6.1.1
Solve 3x + 5 = 20.

Method. Get all the x terms on one side and the numbers on the other, then divide.

  1. Write the equation with the x terms on one side. The x term is already on the left.
    3x + 5 = 20
  2. Move the number terms to the other side. Subtract 5 from both sides, keeping the sign.
    3x = 15
  3. Divide both sides by the number in front of x. Dividing both sides by 3 keeps the equation balanced.
    x = 15 ÷ 3 = 5
  4. Write the answer.
    x = 5

Check. Substitute x = 5 into both sides: 3x + 5 = 20, and 20 = 20. Both sides are equal.

Watch out. Do the same to both sides. Check the sign of each number when you move it across.

Worked example 6.1.2
Solve 4x − 7 = 9.

Method. Get all the x terms on one side and the numbers on the other, then divide.

  1. Write the equation with the x terms on one side. The x term is already on the left.
    4x − 7 = 9
  2. Move the number terms to the other side. Add 7 to both sides.
    4x = 16
  3. Divide both sides by the number in front of x. Dividing both sides by 4 keeps the equation balanced.
    x = 16 ÷ 4 = 4
  4. Write the answer.
    x = 4

Check. Substitute x = 4 into both sides: 4x − 7 = 9, and 9 = 9. Both sides are equal.

Watch out. Do the same to both sides. Check the sign of each number when you move it across.

Worked example 6.1.3
Solve 5x + 3 = 2x + 12.

Method. Get all the x terms on one side and the numbers on the other, then divide.

  1. Collect the x terms on one side. Subtract 2x from both sides so that the x terms are on the left.
    5x + 3 = 2x + 12 becomes 3x + 3 = 12
  2. Move the number terms to the other side. Subtract 3 from both sides, keeping the sign.
    3x = 9
  3. Divide both sides by the number in front of x. Dividing both sides by 3 keeps the equation balanced.
    x = 9 ÷ 3 = 3
  4. Write the answer.
    x = 3

Check. Substitute x = 3 into both sides: 5x + 3 = 18, and 2x + 12 = 18. Both sides are equal.

Watch out. Do the same to both sides. Check the sign of each number when you move it across.

Watch out

Watch out: when you move a number across the equals sign, change its sign. In 4x − 7 = 9, add 7 to both sides to get 4x = 16.

Exam tip

Exam tip. Show every step on its own line. Check your answer in the original equation, not in your working.

Exercise 6A  One-step and two-step equations

Non-calculator
1
FSolve 3x + 5 = 20.
2
FSolve 4x − 7 = 9.
3
FSolve 2x + 3 = -7.
4
FSolve 5x − 2 = 13.
5
FSolve 6x + 1 = -11.
6
FSolve 5x + 3 = 2x + 12.
7
FSolve 4x − 1 = x + 8.
8
FSolve 7x + 2 = 3x − 6.
9
FSolve 3x + 9 = −x + 1.
10
FSolve 6x − 4 = 2x + 8.
11
FSolve 8x + 5 = 3x − 5.
12
FSolve 2x + 9 = 4x + 1.
13
PSAli has 3 bags of x sweets and 5 loose sweets. There are 20 sweets altogether. Write an equation and solve it.
14
RAli says that 2x = 10 has the solution x = 20. Is Ali right? Show how you know.
15
RExplain how to solve 5x − 3 = 2x + 9 in two steps. Then solve it.
6.2

Equations with brackets and fractions

Year 8
Solve equations with brackets, Solve equations with fractions
Video lesson: Equations with brackets and fractionsComing soon
A shop sells pencil cases at £x each. Mia buys 2 cases and a pen that costs £3, and her total is £14. Priya also has an equation with a fraction when she shares the money between two friends.

To solve an equation with a bracket, multiply out the bracket first. For example 2(x + 3) = 14 becomes 2x + 6 = 14.

To solve an equation with a fraction, multiply both sides by the bottom number first. This clears the fraction.

Key ideas

  • Expand the bracket by multiplying every term inside it.
  • Multiply both sides by the bottom number to clear a fraction.
  • x / 4 means the whole of x divided by 4. Multiply both sides by 4 to undo it.
  • Check your answer by substituting.
Worked example 6.2.1
Solve 2(x + 3) = 14.

Method. Expand the bracket first, then solve the equation as usual.

  1. Expand the bracket. Multiply every term inside the bracket by 2.
    2(x + 3) = 2x + 6
  2. Move the number to the other side. Subtract 6 from both sides.
    2x = 8
  3. Divide both sides by the number in front of x. Divide both sides by the same number.
    x = 8 ÷ 2 = 4
  4. Write the answer.
    x = 4

Check. Substitute x = 4: 2 × (7) = 14, which equals 14.

Watch out. Multiply every term in the bracket, including the number. Forgetting the number is the most common mistake.

Worked example 6.2.2
Solve (2x − 1)/3 = 3.

Method. Multiply both sides by the bottom number to remove the fraction.

  1. Multiply both sides by the bottom number. This clears the fraction and keeps the equation balanced.
    (2x − 1) = 3 × 3 = 9
  2. Move the number to the other side. Add 1 to both sides.
    2x = 10
  3. Divide both sides by the number in front of x.
    x = 10 ÷ 2 = 5
  4. Write the answer.
    x = 5

Check. Substitute x = 5: (9) ÷ 3 = 3, which equals 3.

Watch out. Multiply both sides by the bottom number first. Do not multiply just the top of the fraction.

Worked example 6.2.3
Solve x/4 + 3 = 7.

Method. Move the number to the other side, then multiply both sides by the bottom number.

  1. Move the number to the other side. Subtract 3 from both sides.
    x/4 = 4
  2. Multiply both sides by 4. This removes the fraction from x over the bottom number.
    x = 4 × 4
  3. Write the answer.
    x = 16

Check. Substitute x = 16: 16 ÷ 4 + 3 = 7, which equals 7.

Watch out. Multiply by the bottom number, not divide. If x is divided by 4, multiply both sides by 4.

Watch out

Watch out: in 2(x + 3) = 14, the 2 multiplies the 3 as well. It is 2x + 6, not 2x + 3.

Exam tip

Exam tip. Write the equation with the fraction removed as a new line before you solve it. Examiners look for the steps.

Exercise 6B  Equations with brackets and fractions

Non-calculator
1
FSolve 2(x + 3) = 14.
2
FSolve 3(2x − 1) = 15.
3
FSolve 4(x − 2) = 12.
4
FSolve 5(2x + 1) = 35.
5
FSolve -2(x + 4) = 6.
6
FSolve 3(x + 1) = 18.
7
FSolve 2(3x − 2) = 8.
8
FSolve (x + 3)/2 = 5.
9
FSolve (2x − 1)/3 = 3.
10
FSolve (x + 5)/4 = 3.
11
FSolve (3x + 2)/5 = 4.
12
FSolve x/4 + 3 = 7.
13
FSolve x/5 − 2 = 3.
14
FSolve x/3 + 1 = 5.
15
RSolve (2x − 3)/5 = 1.
16
PSMia buys 2 pencil cases for £x each and a pen for £3. She pays £14 in total. Write an equation and solve it.
17
PSSolve 4(x − 2) = 2x + 8.
18
RAli says that (x + 2)/3 = 4 has the solution x = 10. Is Ali right? Show your working.
19
PSSolve (x + 5)/4 = 3. Show each step.
20
PSA number x is divided by 6, then 4 is added. The answer is 9. Write an equation and solve it.
21
PSSolve 3(2x + 1) = 5x + 13.
22
RSolve (3x + 2)/2 = 7 − x.
6.3

Forming and solving equations

Year 8
Form and solve an equation from a context
Video lesson: Forming and solving equationsComing soon
Ben and his sister Ella are in a quiz. Ben has 3 points more than Ella. Their total is 27 points. Ben writes an equation to find out how many points Ella scored.

Many problems can be solved by writing an equation. Follow these steps:

  1. choose a letter for the unknown and say what it means;
  2. write an equation using the information;
  3. solve the equation;
  4. answer the question in words, with units.

Key ideas

  • Choose a letter and say clearly what it stands for.
  • Write the equation from the information in the question.
  • Solve the equation, then answer the question in words.
  • Check that your answer makes sense in the story.
Worked example 6.3.1
Ella scored x points. Ben scored x + 3. Their total is 27. Write and solve an equation.

Method. Get all the x terms on one side and the numbers on the other, then divide.

  1. Write the equation with the x terms on one side. The x term is already on the left.
    2x + 3 = 27
  2. Move the number terms to the other side. Subtract 3 from both sides, keeping the sign.
    2x = 24
  3. Divide both sides by the number in front of x. Dividing both sides by 2 keeps the equation balanced.
    x = 24 ÷ 2 = 12
  4. Write the answer.
    x = 12

Check. Substitute x = 12 into both sides: 2x + 3 = 27, and 27 = 27. Both sides are equal.

Watch out. Do the same to both sides. Check the sign of each number when you move it across.

Worked example 6.3.2
A number is multiplied by 3 and 5 is added. The answer is 26. Find the number.

Method. Get all the x terms on one side and the numbers on the other, then divide.

  1. Write the equation with the x terms on one side. The x term is already on the left.
    3x + 5 = 26
  2. Move the number terms to the other side. Subtract 5 from both sides, keeping the sign.
    3x = 21
  3. Divide both sides by the number in front of x. Dividing both sides by 3 keeps the equation balanced.
    x = 21 ÷ 3 = 7
  4. Write the answer.
    x = 7

Check. Substitute x = 7 into both sides: 3x + 5 = 26, and 26 = 26. Both sides are equal.

Watch out. Do the same to both sides. Check the sign of each number when you move it across.

Worked example 6.3.3
Three consecutive whole numbers add up to 51. Find them.

Method. Get all the x terms on one side and the numbers on the other, then divide.

  1. Write the equation with the x terms on one side. The x term is already on the left.
    3x + 3 = 51
  2. Move the number terms to the other side. Subtract 3 from both sides, keeping the sign.
    3x = 48
  3. Divide both sides by the number in front of x. Dividing both sides by 3 keeps the equation balanced.
    x = 48 ÷ 3 = 16
  4. Write the answer.
    x = 16

Check. Substitute x = 16 into both sides: 3x + 3 = 51, and 51 = 51. Both sides are equal.

Watch out. Do the same to both sides. Check the sign of each number when you move it across.

Watch out

Watch out: the unknown must be defined. Write “Let x be ...” before you write the equation.

Exam tip

Exam tip. Always answer the question in words, with units if there are any. A value of x alone is not the final answer when the question asks for the age or the cost.

Exercise 6C  Forming equations from contexts

Non-calculator
1
FA number is multiplied by 3, and 5 is added. The answer is 26. Let the number be x. Write an equation and solve it.
2
FBen is 4 years older than Ali. The sum of their ages is 26. Let Ali be a years old. Write and solve an equation.
3
PSThree consecutive whole numbers add up to 51. Let the smallest be x. Find the three numbers.
4
PSThe angles of a triangle are x, 2x and 3x. Write an equation and find x. Use the fact that angles in a triangle add up to 180°.
5
PSTwo angles on a straight line are x and 3x − 20. Write an equation and find both angles.
6
PSTwice a number, minus 7, equals the number plus 3. Let the number be x. Find x.
7
PSPens cost £x each. Sam buys 3 pens and pays £2 for a rubber, a total of £14. Write an equation and find the cost of one pen.
8
PSThe length of a rectangle is x + 3 and its width is x. The perimeter is 30 cm. Write an equation and find the length.
9
PSMia thinks of a number, doubles it and takes away 6. The answer is 14. Let the number be x. Find x.
10
PSA taxi charges £3 plus £2 for each mile. The total fare is £15. Let the number of miles be x. How many miles were travelled?
11
RAli says that the equation 3x + 4 = 3x + 7 has the solution x = 1. Is Ali right? Explain how you know.
12
PSA rectangle has a length of 2x and a width of x + 4. Its perimeter is 32 cm. Find the length.
13
RAli says the sum of two consecutive even numbers can be 50. Let the smaller number be x, so the numbers are x and x + 2. Is Ali right?
14
PSPriya has 4 more coins than Ben. Together they have 30 coins. Let Ben have x coins. How many does each have?
6.4

Linear inequalities and number lines

Year 8
Solve linear inequalities in one variable, Show inequalities on a number line, List integer solutions
Video lesson: Linear inequalities and number linesComing soon
A school trip allows at most 40 children on each bus. The teacher writes an inequality to show the number of children, and shows it on a number line.

An inequality uses one of these signs: < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to).

Solve an inequality in the same way as an equation. But if you multiply or divide by a negative number, you must turn the sign round.

On a number line, use a closed circle for ≤ or ≥ (the value is included), and an open circle for < or > (not included).

Key ideas

  • Solve an inequality like an equation, keeping the sign.
  • Dividing by a negative number reverses the sign: −2x < 6 gives x > −3.
  • Closed circle (filled): ≤ or ≥. Open circle: < or >.
  • List integer solutions by counting whole numbers in the range.
Worked example 6.4.1
Solve 3x − 2 < 10.

Method. Solve the inequality as an equation, but keep the inequality sign. Turn the sign round only when you divide by a negative number.

  1. Move the number term to the other side. Add 2 to both sides.
    3x < 12
  2. Divide both sides by the number in front of x. The number in front of x is positive, so the sign stays the same.
    x < 12 ÷ 3 = 4
  3. Write the answer.
    x < 4

Check. Test x = 4 in the equation: 3x − 2 = 10, which is 10 at the boundary.

Watch out. Dividing or multiplying by a negative number reverses the inequality sign.

Worked example 6.4.2
Solve −2x > 6.

Method. Solve the inequality as an equation, but keep the inequality sign. Turn the sign round only when you divide by a negative number.

  1. Move the number term to the other side. Subtract 0 from both sides, keeping the sign.
    −2x > 6
  2. Divide both sides by the number in front of x. If you divide by a negative number, the inequality sign must turn round.
    x < 6 ÷ -2 = −3
  3. Write the answer.
    x < −3

Check. Test x = −3 in the equation: −2x = 6, which is 6 at the boundary.

Watch out. Dividing or multiplying by a negative number reverses the inequality sign.

Worked example 6.4.3
Solve 5 − x ≤ 2.

Method. Solve the inequality as an equation, but keep the inequality sign. Turn the sign round only when you divide by a negative number.

  1. Move the number term to the other side. Subtract 5 from both sides, keeping the sign.
    −x ≤ -3
  2. Divide both sides by the number in front of x. If you divide by a negative number, the inequality sign must turn round.
    x ≥ -3 ÷ -1 = 3
  3. Write the answer.
    x ≥ 3

Check. Test x = 3 in the equation: −x + 5 = 2, which is 2 at the boundary.

Watch out. Dividing or multiplying by a negative number reverses the inequality sign.

Watch out

Watch out: −2x > 6 gives x < −3, not x > −3. The sign turns round when you divide by a negative number.

Exam tip

Exam tip. Check your answer by substituting a value inside the range, and one outside it. The answer should be right for the first and wrong for the second.

Exercise 6D  Solving inequalities and number lines

Non-calculator
1
FSolve 3x − 2 < 10.
2
FSolve 2x + 5 ≥ 1.
3
FSolve −2x > 6.
4
FSolve 5x − 1 ≤ 9.
5
FSolve −x + 5 ≤ 2.
6
FSolve 4x + 1 > -7.
7
FSolve 3x + 4 ≥ -5.
8
FSolve −3x + 2 < 8.
9
PSShow the solution of x > −2 on a number line.
-6-5-4-3-2-10123456
10
PSShow the solution of x ≤ 3 on a number line.
-6-5-4-3-2-10123456
11
FList all the integers that satisfy −2 < x ≤ 4.
12
PSWhich integers satisfy both 3x − 2 < 10 and x ≥ −3? List them.
13
RAli says that if 2x < 6, then x < 3. Is Ali right? Explain how you know.
14
RAli says that if −2x < 6, then x < −3. Is Ali right? Explain how you know.
15
PSSolve 4x + 1 > −7. Show the solution on a number line.
-6-5-4-3-2-10123456
16
PSSolve 3 − 2x ≥ 9.
17
PSA bus holds at most 40 children. There are already 15 children on board, and x more get on. Write an inequality and solve it.
18
PSSolve −3x + 2 < 8.
6.5

Simultaneous equations (basics)

Year 8
Solve two simultaneous linear equations by elimination or substitution
Video lesson: Simultaneous equations (basics)Coming soon
A cinema sells adult and child tickets. Two groups buy tickets: 2 adults and 3 children cost £19, and 1 adult and 2 children cost £11. The manager wants to know the price of each ticket, so she writes two equations.

Two equations with the same two unknowns are called simultaneous equations. Solving them means finding the pair of values that works in both equations.

Elimination: multiply one or both equations so that one unknown has the same number in front of it, then add or subtract the equations to remove it.

Substitution: if one equation says y = (something), replace y with that in the other equation.

Key ideas

  • Simultaneous equations have the same two unknowns in both equations.
  • Elimination: match the number in front of one unknown, then add or subtract.
  • Substitution: put an expression for one unknown into the other equation.
  • Always check your solution in both original equations.
Worked example 6.5.1
Solve x + y = 10 and x − y = 4.

Method. Make the x terms match, then add or subtract the equations to remove x. Solve for y, then find x.

  1. Label the equations. We label them so we can refer to them.
    (1) x + y = 10 and (2) x − y = 4
  2. Multiply to make the x terms match. Multiply equation (1) by 1 and equation (2) by 1.
    (1) × 1: x + y = 10 and (2) × 1: x − y = 4
  3. Subtract to remove x. Subtract the second new equation from the first, so the x terms cancel.
    2y = 6
  4. Solve for y. Divide both sides by the number in front of y.
    y = 3
  5. Substitute y back to find x. Put y into one of the original equations, and solve for x.
    Use (1): x + 1 × (3) = 10, so x = 7
  6. Write the answer.
    x = 7, y = 3

Check. Check both equations: 1 × (7) + 1 × (3) = 10 = 10, and 1 × (7) − 1 × (3) = 4 = 4. ✓

Watch out. Check your answer in both original equations, not just one.

Worked example 6.5.2
Solve y = 2x − 1 and x + y = 8.

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

  1. Write the equations. The first equation already says what y is equal to.
    y = 2x − 1 and x + 1y = 8
  2. Substitute into the second equation. Replace y with 2x − 1.
    x + 1(2x − 1) = 8
  3. Expand and collect the x terms. Multiply the bracket, then collect the x terms.
    3x − 1 = 8
  4. Solve for x. Move the number and divide by the number in front of x.
    x = 3
  5. Substitute x back to find y. Use the first equation.
    y = 2 × (3) − 1 = 5
  6. Write the answer.
    x = 3, y = 5

Check. Check in the second equation: 1 × (3) + 1 × (5) = 8 = 8. ✓

Watch out. Put the expression in brackets before you multiply, so that every term is multiplied.

Worked example 6.5.3
Solve 2x + 3y = 16 and x + 2y = 11.

Method. Make the x terms match, then add or subtract the equations to remove x. Solve for y, then find x.

  1. Label the equations. We label them so we can refer to them.
    (1) 2x + 3y = 16 and (2) x + 2y = 11
  2. Multiply to make the x terms match. Multiply equation (1) by 1 and equation (2) by 2.
    (1) × 1: 2x + 3y = 16 and (2) × 2: 2x + 4y = 22
  3. Subtract to remove x. Subtract the second new equation from the first, so the x terms cancel.
    −y = -6
  4. Solve for y. Divide both sides by the number in front of y.
    y = 6
  5. Substitute y back to find x. Put y into one of the original equations, and solve for x.
    Use (1): 2x + 3 × (6) = 16, so x = −1
  6. Write the answer.
    x = −1, y = 6

Check. Check both equations: 2 × (−1) + 3 × (6) = 16 = 16, and 1 × (−1) + 2 × (6) = 11 = 11. ✓

Watch out. Check your answer in both original equations, not just one.

Watch out

Watch out: when you subtract equations, subtract every term in the second equation. Changing the signs is the most common error.

Exam tip

Exam tip. Write the equations as (1) and (2). Write each step on its own line, so the method is clear even if a number is wrong.

Exercise 6E  Simultaneous equations

Non-calculator
1
FSolve the simultaneous equations x + y = 10 and x − y = 4.
2
FSolve the simultaneous equations 2x + y = 9 and x + y = 7.
3
FSolve the simultaneous equations 3x + 2y = 14 and x + 2y = 6.
4
FSolve the simultaneous equations x + 3y = 19 and 2x + y = 8.
5
FSolve the simultaneous equations 2x + 3y = 16 and 4x − y = 18.
6
FSolve the simultaneous equations x + 4y = -5 and 2x + y = 4.
7
FSolve the simultaneous equations 2x + 3y = 24 and 3x − 2y = 10.
8
FSolve the simultaneous equations 2x + y = 1 and x + 2y = 5.
9
FSolve y = 3x + 2 and 2x + y = 12.
10
FSolve y = x + 4 and 3x + y = 12.
11
PSTwo adults and three children pay £19. One adult and two children pay £11. Let the adult ticket cost x and the child ticket cost y. Find both prices.
12
PSThree pens and two rulers cost £8. One pen and two rulers cost £4. Let a pen cost x and a ruler cost y. Find the cost of a pen and of a ruler.
13
RSam says that x = 2 and y = 3 is a solution of x + y = 5 and 2x − y = 1. Is Sam right? Check both equations.
14
PSSolve 2x + y = 1 and x + 2y = 5.
15
PSSolve x + 3y = 19 and 2x + y = 8.
16
RSolve y = 4 − x and 2x + 3y = 13.
17
RSolve 4x − y = 18 and 2x + 3y = 16. Explain which method you chose and why.

Chapter summary

SkillMethod
Solve a linear equationUndo the sums first, then divide: collect the x terms on one side
Brackets and fractionsExpand the bracket, or multiply both sides by the bottom number, then solve
Forming equationsDefine the unknown, write the equation from the story, solve, then answer in words
InequalitiesSolve like an equation; turn the sign round when you divide by a negative number
Number lineOpen circle for < or >; closed circle for ≤ or ≥
Simultaneous equationsElimination: match one unknown and add or subtract. Substitution: replace one unknown
CheckSubstitute your answers into the original equations

Can you…?

  • I can solve equations with the unknown on one side or both sides
  • I can solve equations with brackets and fractions
  • I can write an equation from a context and solve it
  • I can solve an inequality, and turn the sign round when I divide by a negative number
  • I can show an inequality on a number line, with the right type of circle
  • I can solve simultaneous equations by elimination or substitution, and check both equations

Key terms

equation
two expressions joined by an equals sign, which can be solved to find the unknown
inequality
two expressions joined by a sign such as <, >, ≤ or ≥
solution
the value (or values) that make the equation or inequality true
open circle
used on a number line when the value is not included (< or >)
closed circle
used on a number line when the value is included (≤ or ≥)
integer
a whole number, which can be positive, negative or zero
simultaneous equations
two equations with the same two unknowns, solved together

Exercise 6Z  Check-up

Non-calculator
1
Solve 4x − 9 = 11.
2
Solve 6x + 2 = 3x + 14.
3
Solve 3(x + 4) = 27.
4
Solve (x + 7)/2 = 6.
5
Solve x/3 − 2 = 4.
6
Solve 5x − 2 < 13.
7
List the integers that satisfy −1 < x < 4.
8
Solve y = 2x + 1 and x + y = 10.
9
Solve x + y = 9 and x − y = 1.
10
A number is doubled and then 7 is added. The answer is 25. Find the number.

Total 30 marks. Allow about 30 minutes.

Exercise 6X  Exam-style questions

Non-calculator
1
FSolve 7x + 4 = 3x + 20.[2 marks]
2
FSolve 5(2x − 1) = 45.[3 marks]
3
FSolve (3x + 1)/4 = 5.[2 marks]
4
FSolve 6 − 2x ≥ 10.[3 marks]
5
FSolve 2x + 3 < 11 and show the solution on a number line.[3 marks]
-6-5-4-3-2-10123456
6
FList the integers that satisfy both x > −3 and x ≤ 2.[2 marks]
7
FSolve x + 3y = 11 and 2x − y = 1.[4 marks]
8
FThe length of a rectangle is 3x cm and its width is x + 2 cm. The perimeter is 28 cm. Write an equation and find the length.[4 marks]
9
FAli says that the inequality 2 − x > 5 has the solution x > −3. Is he right? Explain your answer.[3 marks]
10
FSam buys 2 pens and 1 ruler for £7. Lee buys 1 pen and 3 rulers for £11. Let a pen cost x and a ruler cost y. Find the cost of one pen.[4 marks]
← Chapter 5: Expanding and factorisingChapter 7: Sequences and linear graphs →

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