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Forming And Solving Equations

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Notes

Forming Equations from Words

  • Use **x** to represent an unknown value; translate phrases like '2 less than something' → **x − 2**, 'double something' → **2x**, '5 lots of something' → **5x**, '3 more than something' → **x + 3**, 'half of something' → **x/2**.
  • Key words: **sum/total/more than** (+), **difference/less than** (−), **product/lots of/times as many** (×), **shared/split/grouped** (÷).
  • Use brackets to keep order correct: 'something add 1, then multiplied by 3' → **3(x + 1)**; compare with 'something multiplied by 3, then add 1' → **3x + 1**.
  • Choose the unknown wisely: if Adam is 10 years younger than Barry, let Barry's age=age = **x**, Adams=Adam's = **x − 10**; if Adam's age is half Barry's, let Barrys=Barry's = **2x**, Adams=Adam's = **x**.
  • An **equation** has an equals sign; insert 'is equal to' to place it. E.g., 'Lisa's age is double Aisha's and sum is 27' → **2x +x=+ x = 27**.
  • Always give the answer **in context** (e.g., 'Lisa is 18 years old').

Forming Equations from Shapes

  • Use all given information and **shape properties**: triangles (equilateral, isosceles, scalene, right-angled), quadrilaterals (square, rectangle, kite, rhombus, parallelogram, trapezium), polygons (sum of interior angles = **180(n − 2)**), parallel lines (alternate, corresponding, co-interior).
  • For **perimeter**, add all side lengths; for **area**, use appropriate formula (e.g., rectangle: length × width, triangle: ½ × base ×height)\times height).
  • For **3D shapes**, volume = cross‑section area × length (prisms); surface area =sum= sum of face areas.
  • Sketch a diagram if none given; split irregular shapes into sum/difference of common shapes.
  • Put **brackets** around algebraic expressions when substituting into formulas (e.g., perimeter =2(3x+1)+2(2x5))= 2(3x + 1) + 2(2x - 5)).
  • Read carefully: does the question ask for an angle, perimeter, area, or something else?

Problem Solving with Equations

  • Problem solving involves forming and solving equations from a **real‑life or constructed situation**; equations can be **linear** (e.g.,2(x+4)=3x)(e.g., 2(x + 4) = 3x) or **quadratic** (e.g.,x27x+12=0)(e.g., x^{2} - 7x + 12 = 0).
  • Answers must be given **in context** with correct units (e.g., 'The population density is 225 people per square km').
  • To solve quadratic equations, bring all terms to one side to get '= 0'; choose method (factorising, quadratic formula, completing square).
  • If two solutions arise, **justify** which is correct (e.g., length cannot be negative).
  • Use algebra in other settings: percentages (P%=P100)(P\% = \frac{P}{100}), ratios (e.g., x : (x+2)=5(x + 2) = 5 : 8 → x/(x+2)=58)x/(x+2) = \frac{5}{8}), or unfamiliar equations (e.g.,12x=7x(e.g., \frac{12}{x} = 7 - x → multiply by x to get quadratic).
  • If part (a) asks to prove an equation and part (b) uses it, you can still do part (b) without having done part (a).

Example: Forming Linear Equations from Words

  • A flowerbed has red, yellow, and purple flowers. Yellow =3×= 3 \times red; purple = yellow + 5; difference (purplered)=29(purple - red) = 29.
  • Let red=red = **x** → yellow = **3x**, purple = **3x + 5**; equation: (3x+5)x=29(3x + 5) - x = 292x+5=292x + 5 = 29x=12x = 12.
  • Answer: yellow flowers =3×12== 3 \times 12 = **36**.

Example: Forming Equations from Shapes

  • Rectangle length =(3x+1)= (3x + 1) cm, width =(2x5)= (2x - 5) cm, perimeter =22= 22 cm.
  • Perimeter =2(3x+1)+2(2x5)=6x+2+4x10=10x= 2(3x + 1) + 2(2x - 5) = 6x + 2 + 4x - 10 = 10x - 8; set equal to 22 → 10x8=2210x - 8 = 22x=3x = 3.
  • Area = length × width =(3×3+1)×(2×35)=10×1== (3\times 3 + 1) \times (2\times 3 - 5) = 10 \times 1 = **10 cm²**.

Example: Problem Solving with a Cube Net

  • Cube net has 14 edges; perimeter = **14x** cm. Net has 6 faces; area = **6x²** cm2cm^{2}.
  • Volume of cube = **x³** cm3cm^{3}. Difference (volume − surface area)=8×area) = 8 \times perimeter → x36x2=112xx^{3} - 6x^{2} = 112xx36x2112x=0x^{3} - 6x^{2} - 112x = 0.
  • Factor out x(x0):x(x26x112)=0x (x \ne 0): x(x^{2} - 6x - 112) = 0x26x112=0x^{2} - 6x - 112 = 0. Solve: x=14x = 14 or x=8(reject)x = -8 (reject). Volume =143== 14^{3} = **2744 cm³**.

Common Pitfalls & Tips

  • Always **read the question** to see what is asked (angle, perimeter, area, etc.).
  • For surface area/volume, check the formula sheet provided in the exam.
  • When forming equations, **double‑check** that the algebraic expression matches the wording (e.g., '5 less than x' is x − 5, not 5 − x).
  • If you get a quadratic, ensure you have =0'= 0' before solving; discard extraneous solutions based on context.

Translating Words to Expressions

Words to Expressions'2 less than something' → x − 2'double something' → 2x'5 lots of something' → 5x'3 more than something' → x + 3'half of something' → x/2Key words:+ : sum, total, more than, increase− : difference, less than, decrease× : product, lots of, times as many÷ : shared, split, grouped, halved

Forming Equation from Rectangle Perimeter

Rectangle: Perimeter = 22 cm3x + 12x − 5Perimeter = 2(3x+1) + 2(2x−5) = 6x+2 + 4x−10 = 10x−8Set equal to 22: 10x−8 = 22Solve: 10x = 30 → x = 3Area = (3×3+1)×(2×3−5) = 10×1 = 10 cm²

Cube Net and Equations

Cube Net (side length x)Perimeter of net = 14x cmArea of net = 6x² cm²Volume of cube = x³ cm³Given: x³ − 6x² = 8 × 14x→ x² − 6x − 112 = 0 → x = 14

Triangle with Equal Base Angles

Isosceles Triangle ABCABCAB = 3x−5AC = 19−xBC = 2x∠ABC = ∠BCA (base angles equal)→ AB = AC → 3x−5 = 19−xSolve: 4x = 24 → x = 6Perimeter = (3×6−5)+(19−6)+2×6 = 13+13+12 = 38 cm

Practice questions

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  1. 1.Which expression represents '5 less than twice a number x'?

    Easy
    • A2x52x - 5
    • B52x5 - 2x
    • C2(x5)2(x - 5)
    • D2x+52x + 5
  2. 2.The perimeter of a rectangle is 30 cm. Its length is 9 cm. Which equation finds its width w?

    Easy
    • A2(9+w)=302(9 + w) = 30
    • B9+w=309 + w = 30
    • C2(9)+w=302(9) + w = 30
    • D9w=309w = 30
  3. 3.The sum of three consecutive integers is 36. If the smallest is x, which equation represents this?

    Easy
    • Ax+(x+1)+(x+2)=36x + (x+1) + (x+2) = 36
    • Bx+(x+1)+(x+3)=36x + (x+1) + (x+3) = 36
    • Cx+(x+2)+(x+4)=36x + (x+2) + (x+4) = 36
    • D3x=363x = 36
  4. 4.A triangle has angles x°, (x+10)° and (x+20)°. Find x.

    Medium
    • A50
    • B60
    • C70
    • D40
  5. 5.The length of a rectangle is 3 cm more than its width. Its area is 40cm240 cm^{2}. If the width is w cm, which equation finds w?

    Medium
    • Aw(w+3)=40w(w+3) = 40
    • Bw+(w+3)=40w + (w+3) = 40
    • C2w+2(w+3)=402w + 2(w+3) = 40
    • Dw(w3)=40w(w-3) = 40
  6. 6.In the diagram, K, L and M lie on a circle, centre O. Angle KML=2xKML = 2x^{\circ} and reflex angle KOL=11xKOL = 11x^{\circ}. Find x.

    Medium
    • A15
    • B20
    • C25
    • D30
  7. 7.Julie multiplies a number by 5 and adds 4. Liam subtracts the number from 10. Julie's answer is two thirds of Liam's answer. Find the number.

    Medium
    • A1
    • B2
    • C3
    • D4
  8. 8.The diagram shows a trapezium. All measurements are in metres. The area of the trapezium is 138m2138 m^{2}. Work out the value of x.

    Hard
    • A6
    • B7
    • C8
    • D9

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