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Algebra - Finding missing numbers in equations

Grade 5IGCSE

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

🔑Concepts

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Inverse Operations: Using the opposite operation to isolate the missing number (e.g., addition is the inverse of subtraction).

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Equation Balance: An equation is like a balance scale; whatever operation you perform on one side, you must perform on the other to keep it equal.

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Variables: Using letters like xx, yy, or nn (or symbols like □\square) to represent an unknown value.

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Substitution: Checking your answer by plugging the result back into the original equation to see if it makes sense.

📐Formulae

If x+a=bx + a = b, then x=b−ax = b - a

If x−a=bx - a = b, then x=b+ax = b + a

If a×x=ba \times x = b, then x=b÷ax = b \div a

If x÷a=bx \div a = b, then x=b×ax = b \times a

💡Examples

Problem 1:

Find the value of nn in the equation: n+15=42n + 15 = 42.

Solution:

n=27n = 27

Explanation:

To find nn, use the inverse of addition. Subtract 15 from both sides: 42−15=2742 - 15 = 27.

Problem 2:

Solve for xx: x−12=35x - 12 = 35.

Solution:

x=47x = 47

Explanation:

To isolate xx, use the inverse of subtraction. Add 12 to both sides: 35+12=4735 + 12 = 47.

Problem 3:

Find the missing number: 6×y=546 \times y = 54.

Solution:

y=9y = 9

Explanation:

To find yy, use the inverse of multiplication. Divide both sides by 6: 54÷6=954 \div 6 = 9.

Problem 4:

Solve the equation: z÷4=12z \div 4 = 12.

Solution:

z=48z = 48

Explanation:

To find zz, use the inverse of division. Multiply both sides by 4: 12×4=4812 \times 4 = 48.

Problem 5:

A harder one: 50−x=1850 - x = 18. What is xx?

Solution:

x=32x = 32

Explanation:

In this case, we can swap xx and 18, or think: '50 minus what gives 18?' By calculating 50−1850 - 18, we find x=32x = 32.