Year 9 · Unit 8 Ready for Year 10 · Lesson 6 of 6

Three Year 9 ideas, one Year 10 problem

Index laws, Pythagoras' theorem and linear equations are the Year 9 tools that Year 10 uses together. Bring each one back, then join them with a squared bracket to solve a triangle that looks harder than it is.

Number and AlgebraNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From lesson 3: expand (x − 4)².
    Answerx² − 8x + 16.
  2. From Unit 1: write 10−2 as a fraction.
    Answer1/100: a negative index means one over the positive power.
  3. From Unit 7: solve 5x + 3 = 2x + 18.
    Answerx = 5: 3x = 15.
Words used in this lesson
hypotenuse
the longest side of a right-angled triangle, opposite the right angle. For example: in a 3, 4, 5 triangle it is 5.
index
the small raised number that says how many times to multiply. For example: in 25 the index is 5.
linear equation
an equation with x but no x² or higher power. For example: 4x + 1 = 13.
cancel
remove the same term from both sides. For example: x² + 7 = x² + x becomes 7 = x.
retrieval
bringing back something learned earlier. For example: the warm-up questions.

A triangle with x in it

A fire service ladder stands on level ground and leans against a wall. Its foot is x metres from the wall, its top reaches 12 metres up the wall, and the ladder is 8 metres longer than x. The ground, the wall and the ladder make a right-angled triangle, so Pythagoras says x² + 12² = (x + 8)².

Predict first: is this a Year 10 quadratic equation, or can Year 9 algebra solve it? Then try the figure, which has four triangles of the same kind (not this one).

x² + 6² = (x + 2)²
3
45x² + k², the two shorter sides squared
25(x + m)², the rod squared
20how far apart the two are

x² + 6² = (x + 2)² at x = 3: 45 against 25. The rod is too short.

Show me the steps
  1. Shorter sides x and 6, rod x + 2, at x = 3: x² + 36 = 45 but (x + 2)² = 25. The rod is too short.
  2. Expand: x² + 36 = x² + 4x + 4. Take x² from both sides: 36 = 4x + 4, so 4x = 32 and x = 8. At x = 8: 64 + 36 = 100 = 10².
  3. Shorter sides x and 20, rod x + 10: 400 = 20x + 100, so x = 15, and 225 + 400 = 625 = 25².
  4. Shorter sides x and 24, rod x + 18: 576 = 36x + 324, so x = 7. Shorter sides x and 9, rod x + 1: 81 = 2x + 1, so x = 40.

Year 10 algebra is built from Year 9 pieces. Here Pythagoras gives the equation, expanding a square (this unit) turns (x + 8)² into x² + 16x + 64, the index law x × x = x² says what each cell is, and balancing (Unit 7) finishes it.

For the ladder: x² + 144 = x² + 16x + 64. Take x² from both sides: 144 = 16x + 64, so 16x = 80 and x = 5. The foot is 5 m from the wall and the ladder is 13 m long: 25 + 144 = 169 = 13². Year 9 algebra was enough, because the x² terms cancel.

A common mistake: taking the square root of each term. The square root of x² + 144 is not x + 12: at x = 5 it is 13, not 17.

Bring it back: three Year 9 ideas

For each idea there is a worked example, then one with the first step given, then questions to do alone below.

Index laws. Worked: simplify (x4)² ÷ x³.

  1. A power of a power multiplies the indices: (x4)² = x8.
  2. Dividing subtracts them: x5.

Index laws. Your go, first step given: work out 3² × 34 ÷ 3³.

  1. Multiplying adds the indices: 3² × 34 = 36.
  2. Now divide by 3³ and find the value.
Check the answer

36 ÷ 3³ = 3³ = 27.

Linear equations. Worked: solve 5(x − 2) = 3x + 6.

  1. Expand: 5x − 10 = 3x + 6.
  2. Take 3x from both sides and add 10: 2x = 16, so x = 8.
  3. Check: 5 × 6 = 30 and 24 + 6 = 30.

Linear equations. Your go, first step given: solve 2(3x + 1) = 4x + 16.

  1. Expand: 6x + 2 = 4x + 16.
  2. Now gather the x terms and the numbers, and check.
Check the answer

2x = 14, so x = 7. Check: 2 × 22 = 44 and 28 + 16 = 44.

Pythagoras with a bracket. Your go, first step given: a right-angled triangle has shorter sides x and 10 and longest side x + 2. Find x.

  1. Pythagoras and expand: x² + 100 = x² + 4x + 4.
  2. Now cancel, solve and check the three sides.
Check the answer

100 = 4x + 4, so 4x = 96 and x = 24. The sides are 24, 10 and 26: 576 + 100 = 676 = 26².

Check before you go on

1. Which equation becomes a linear equation once you expand and simplify?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Work out (2³)² ÷ 24.

3. A right-angled triangle has shorter sides of 9 cm and 12 cm. How long is the longest side, in centimetres?

Level 2 · Building

4. Solve 3(x − 4) = x + 10.

5. Solve (x + 3)² = x² + 33.

Level 3 · Stretch

6. A right-angled triangle has shorter sides x cm and 21 cm. Its longest side is (x + 7) cm. Find the perimeter of the triangle, in centimetres.

7. (x³ + 2)(x³ − 2) = xn − 4. What is n?

Something to think about

Try a triangle with shorter sides x and 4 and longest side x + 4. What does the algebra give, and why does it make sense?

One way to think about it

x² + 16 = x² + 8x + 16 gives 8x = 0, so x = 0: a triangle with no width. If the longest side is x + m and the other short side is k, then x = (k² − m²) ÷ 2m, which is positive only when k is bigger than m. Try k = 8 and m = 2 to get a whole-number triangle (x = 15: 15, 8, 17).

Exit ticket

8. Solve (x − 3)² = (x + 1)(x − 5).