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Algebra - Algebraic Representation and Formulae

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Translating verbal phrases into algebraic expressions (e.g., '5 less than x' becomes x−5x - 5)

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Substitution: Replacing variables with specific numerical values to evaluate an expression or formula

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Changing the subject of a formula: Using inverse operations to isolate a specific variable on one side of the equation

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Constructing formulae: Creating a mathematical relationship based on a real-world scenario or geometric property

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Handling square roots and powers when rearranging: Understanding that the inverse of x2x^2 is ±x\pm\sqrt{x}

📐Formulae

Direct Variation: y=kxy = kx (where kk is a constant)

Subject of Formula (Linear): y=mx+c  ⟹  x=y−cmy = mx + c \implies x = \frac{y - c}{m}

Area of a Trapezium: A=12(a+b)hA = \frac{1}{2}(a + b)h

Average Speed: s=dts = \frac{d}{t}

Compound Interest: A=P(1+r100)nA = P(1 + \frac{r}{100})^n

💡Examples

Problem 1:

Given the formula v2=u2+2asv^2 = u^2 + 2as, calculate vv when u=5u = 5, a=2a = 2, and s=6s = 6.

Solution:

v=52+2(2)(6)=25+24=49=7v = \sqrt{5^2 + 2(2)(6)} = \sqrt{25 + 24} = \sqrt{49} = 7

Explanation:

Substitute the given values into the equation, simplify the terms under the square root, and then take the square root of the result.

Problem 2:

Make xx the subject of the formula: y=3x−54y = \frac{3x - 5}{4}.

Solution:

4y=3x−5  ⟹  4y+5=3x  ⟹  x=4y+534y = 3x - 5 \implies 4y + 5 = 3x \implies x = \frac{4y + 5}{3}

Explanation:

First, multiply both sides by 4 to remove the denominator. Then, add 5 to both sides to isolate the term with xx. Finally, divide by 3.

Problem 3:

A taxi company charges a fixed 'call-out' fee of 5andthen5 and then 2 for every kilometer (kk) traveled. Write a formula for the total cost CC.

Solution:

C=5+2kC = 5 + 2k

Explanation:

The fixed fee is a constant (5), and the variable cost depends on the distance (2×k2 \times k). The sum represents the total cost.

Problem 4:

Rearrange A=4πr2A = 4\pi r^2 to make rr the subject.

Solution:

r2=A4π  ⟹  r=A4πr^2 = \frac{A}{4\pi} \implies r = \sqrt{\frac{A}{4\pi}}

Explanation:

Divide both sides by 4π4\pi to isolate r2r^2, then take the square root of both sides to solve for rr.