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Algebra - Expressions, formulae and equations

Grade 7IGCSE

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

🔑Concepts

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Variables and Terms: Understanding that letters (like x, y, n) represent unknown numbers or quantities.

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Expressions vs. Equations: An expression is a mathematical phrase (e.g., 3x + 5), while an equation contains an equals sign (e.g., 3x + 5 = 11).

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Collecting Like Terms: Simplifying expressions by adding or subtracting terms with the same variable and power.

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Expanding Brackets: Using the distributive law to multiply the term outside the bracket by every term inside.

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Substitution: Replacing variables in an expression or formula with specific numerical values to calculate a result.

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Solving Equations: Using inverse operations (balancing) to isolate the variable on one side of the equation.

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Constructing Formulae: Writing a mathematical rule using variables based on a real-world context.

📐Formulae

a(b+c)=ab+aca(b + c) = ab + ac (Distributive Law)

P=2l+2wP = 2l + 2w (Perimeter of a rectangle)

A=bhA = bh (Area of a parallelogram)

v=u+atv = u + at (Common substitution formula for velocity)

x=y−cmx = \frac{y - c}{m} (Example of rearranging a simple linear formula)

💡Examples

Problem 1:

Simplify the expression: 5x+3y−2x+7y+45x + 3y - 2x + 7y + 4

Solution:

3x+10y+43x + 10y + 4

Explanation:

Group the 'x' terms together (5x−2x=3x5x - 2x = 3x) and the 'y' terms together (3y+7y=10y3y + 7y = 10y). The constant (+4) remains as it is because it has no like terms.

Problem 2:

Expand and simplify: 3(2x−5)+4x3(2x - 5) + 4x

Solution:

10x−1510x - 15

Explanation:

First, multiply the 3 by both terms inside the bracket: 3×2x=6x3 \times 2x = 6x and 3×−5=−153 \times -5 = -15. This gives 6x−15+4x6x - 15 + 4x. Then, collect the like terms: 6x+4x=10x6x + 4x = 10x.

Problem 3:

If a=5a = 5 and b=−2b = -2, find the value of 3a+b23a + b^2.

Solution:

19

Explanation:

Substitute the values into the expression: 3(5)+(−2)23(5) + (-2)^2. Calculate the parts: 15+4=1915 + 4 = 19. Remember that a negative number squared becomes positive.

Problem 4:

Solve the equation: 4x−7=134x - 7 = 13

Solution:

x=5x = 5

Explanation:

Add 7 to both sides to isolate the term with x: 4x=204x = 20. Then, divide both sides by 4 to find x: x=20/4=5x = 20 / 4 = 5.

Problem 5:

Write a formula for the total cost (CC) of nn notebooks at 3eachand3 each and ppensatpens at1.50 each.

Solution:

C=3n+1.5pC = 3n + 1.5p

Explanation:

Multiply the quantity of notebooks by their price (3×n3 \times n) and the quantity of pens by their price (1.5×p1.5 \times p), then add them together to get the total cost.