Finding the Unknown | CBSE Class 7 Maths Notes

This note covers unknown quantities, equations, balanced scales, matchstick and tile patterns, trial and error, inverse operations, systematic solving, brackets, word problems, number tricks, common mistakes, and the history of algebra.

What is an equation, and what does solving it mean?

An unknown quantity is a number or amount whose value we want to find. A letter-number is a letter used to represent a number. For example, let e represent the unknown weight of one fried egg in a balanced-scale problem.

An algebraic expression combines numbers and letter-numbers using arithmetic operations. In the expression 2e, the 2 multiplies e: 2e means 2 × e. The signs +, −, × and ÷ mean addition, subtraction, multiplication and division respectively.

Definition: An equation is a statement that two algebraic expressions are equal. The equals sign, =, separates the two expressions and indicates that they have the same value.

The expression before the equals sign is the Left Hand Side (LHS). The expression after it is the Right Hand Side (RHS). Solving an equation means finding the value or values of its unknowns for which LHS and RHS are equal.

How does a balanced scale represent equality?

Three bread slices hang on the left and two fried eggs on the right of a balanced bar. Each bread slice has weight 2; the egg weight is unknown. The numbers use the same unit of weight.

See Fig. 7.6 in your NCERT textbook

Worked example 1. Three bread slices, each weighing 2 units, balance two equally heavy fried eggs. Let e be the weight of one egg in these units.

Answer: The slices weigh 2 + 2 + 2 = 6. Therefore, e + e = 6, or 2e = 6. Dividing both sides by 2 gives e = 3. Two eggs then weigh 6 units, matching the bread.

A solution must make the original equality true. Naming the unknown and writing an equation connect the picture to a calculation. The letter stands for the weight of one egg, while 2e stands for the combined weight of two eggs.

How do patterns help us form equations?

A sequence is an ordered collection of arrangements or numbers. A position number tells us where an arrangement occurs. In the matchstick sequence, let n represent the position number. The number of sticks at that position is 2n + 1. Position numbers are positive whole numbers, meaning counting numbers with no fractional part.

The first position needs 2 × 1 + 1 = 3 sticks, the second needs 2 × 2 + 1 = 5, and the third needs 2 × 3 + 1 = 7. The expression describes the pattern; setting it equal to a required total gives an equation.

How can Jasmine find the position with 99 sticks?

Worked example 2. Jasmine wants an arrangement containing exactly 99 sticks in the sequence whose position n uses 2n + 1 sticks.

Answer: Form 2n + 1 = 99. Subtract 1 from both sides to obtain 2n = 98. Divide both sides by 2 to obtain n = 49. Checking gives 2 × 49 + 1 = 99, so the required arrangement is at position 49.

Trial and error means substituting different possible values and checking whether the sides become equal. Substitution means replacing a letter-number by a chosen number. The following trials approach Jasmine’s required total.

Trial and error can be inefficient. Operations that undo the calculation give a more systematic route. The situation also matters: exactly 200 sticks would require n = 199/2, where the slash denotes division. This is not a whole-number position, so that arrangement is not possible.

How is Ranjana’s tile pattern different?

Let k be the step number in Ranjana’s tile sequence. The number of tiles is 3k + 1. One way to count is k + k + k + 1; another is k + (2k + 1). Both count the same arrangement.

The first three T-shaped arrangements contain 4, 7 and 10 square tiles. Their horizontal rows contain 3, 5 and 7 tiles, with 1, 2 and 3 further tiles extending downwards from the middle.

Reference: NCERT Class 7, page 174

Worked example 3. Can Ranjana make an arrangement using 100 tiles when step k needs 3k + 1 tiles?

Answer: Solve 3k + 1 = 100. Subtracting 1 gives 3k = 99; dividing by 3 gives k = 33. The check, 3 × 33 + 1 = 100, confirms that step 33 uses exactly 100 tiles.

Which properties let us change an equation without losing equality?

A balanced scale stays balanced when equal weights are removed from its two plates. Similarly, performing the same arithmetic operation on both sides of an equation preserves equality. Choose operations that simplify the expression containing the unknown.

Inverse operations undo one another. Addition and subtraction are inverse operations; multiplication and division are inverse operations. A term is a part added or subtracted in an expression. A factor is a number or expression being multiplied.

Property: Removing an added or subtracted term

The additive inverse of a number is the number that adds to it to give zero. When an added or subtracted term is removed from one side, its additive inverse appears on the other side. This is a shortened way of performing the same operation on both sides.

For an unknown y, 2y + 7 = 21 becomes 2y = 21 − 7. Subtracting 7 from each side removes the added 7 on the left. The shorter working omits the cancelling pair, but the equality is still being preserved.

Property: Removing a factor

If one side is a product, meaning the result of multiplication, removing a non-zero factor, meaning a factor that is not zero, requires dividing the other side by that factor. Thus 2y = 14 becomes y = 14 ÷ 2. Division by zero is not permitted.

Property: Removing a divisor

A quotient is the result of division, and the divisor is the number by which we divide. Let u be an unknown in u/15 = 6. Multiplying both sides by the divisor 15 gives u = 6 × 15 = 90.

These properties also help evaluate a long expression when a related value is known. If 14593 − 1459 + 145 − 14 + 88 = 13353, removing the added 88 means subtracting 88 from the right side. The earlier expression need not be recalculated term by term.

The reason for each change matters more than the appearance of a symbol moving across the equals sign. Identify whether the part being removed is an added term, a factor or a divisor before choosing its inverse operation.

How do we solve an equation with the unknown on one side?

The aim is to leave the unknown alone on one side. First remove an added or subtracted term when that simplifies the expression. Then remove a multiplying factor by division. Record enough working to show the operation applied to both sides.

How can a solution be a fraction?

A fraction represents one number divided by another non-zero number. An equation can have a fractional solution even when all the numbers originally written in it are whole numbers.

Worked example 4. Solve 5x − 4 = 7, where x represents the unknown number.

Answer: Add 4 to both sides: 5x − 4 + 4 = 7 + 4. This gives 5x = 11. Divide both sides by 5 to get x = 11/5. A fractional value is a valid solution when it makes the equation true.

Check by substituting 11/5 into the original equation. Its LHS becomes 5 × (11/5) − 4 = 11 − 4 = 7. This equals the RHS, so the solution is correct. The check tests the starting equation, not merely the last line of working.

What if the added term is negative?

Worked example 5. Solve 11y + (−5) = 61, where y is the unknown.

Answer: Subtract −5 from both sides to remove the term −5. Thus 11y = 61 − (−5) = 66. Dividing by 11 gives y = 6. Checking gives 11 × 6 + (−5) = 66 − 5 = 61.

A negative number is a number less than zero. Subtracting a negative number here has the same effect as adding its positive counterpart. Keep the brackets around −5 while writing the operation, so that the subtraction sign and the negative sign are both visible.

1. Identify the unknown and read the whole expression containing it.
2. Remove the added or subtracted term using the inverse operation.
3. Divide by the remaining non-zero multiplying factor to find the unknown.
4. Substitute the result into the original equation and compare both sides.

The full and shortened methods express the same reasoning. For example, writing 11y = 61 − (−5) abbreviates subtracting −5 from both sides. It does not replace the need to understand why that operation keeps the equation balanced.

How do we solve equations with unknown terms on both sides?

When the same unknown occurs on both sides, first bring the unknown terms to one side through an operation on both sides. A like term has the same letter part; for instance, 6y and 4y can be combined by subtracting their multiplying numbers.

How can we collect the unknown terms?

Worked example 6. Solve 6y + 7 = 4y + 21, where y is the unknown number.

Answer: Subtract 4y from both sides to get 2y + 7 = 21. Subtract 7 to obtain 2y = 14. Divide by 2 to obtain y = 7. The original sides then give 6 × 7 + 7 = 49 and 4 × 7 + 21 = 49.

The operation of subtracting 4y is applied to both sides, even though y is initially unknown. Each occurrence of y represents the same value within the equation, so the subtracted quantities are equal. After simplifying, the equation has the familiar form with an unknown on one side.

Can zero be a solution?

Let s be the unknown in 5s = 3s. Subtracting 3s gives 2s = 0, and dividing by 2 gives s = 0. Checking the original equation gives zero on both sides. We must not divide by s here, because its solution is zero.

Can an equation have no solution?

The hint that four more than a number cannot equal five more than that same number leads to x + 4 = x + 5, with x representing the number. Subtracting x would require 4 = 5, which is false. There is no value of x that can satisfy it.

These cases show why checking and interpreting the result are necessary. Solving does not mean assuming that every equation produces a positive whole number. The operations may give zero, a fraction, or a contradiction that shows no solution is possible.

How should we handle brackets and choose a solving method?

Brackets group an expression so that an operation applies to the whole group. In 28(x + 4), where x is an unknown number, 28 multiplies the entire sum x + 4. This makes it possible to solve either by undoing operations or by opening brackets.

How can we undo operations in reverse order?

Worked example 7. Solve 28(x + 4) + 300 = 1000.

Answer: Subtract 300 to obtain 28(x + 4) = 700. Divide by 28 to obtain x + 4 = 25. Subtract 4 to get x = 21. Checking gives 28 × (21 + 4) + 300 = 700 + 300 = 1000.

Here, the outside addition is undone first. Next, division removes the factor multiplying the bracket. Finally, subtraction removes the number added inside the bracket. Each step leaves a simpler expression while maintaining equality.

What changes if we open the brackets first?

The distributive property means multiplying each term inside the bracket by the outside factor. Thus 28(x + 4) becomes 28x + 112. Combining the known numbers gives 28x + 412 = 1000. Subtracting 412 gives 28x = 588, and dividing by 28 gives x = 21.

Another route divides the entire original equation by 4. Since 28, 300 and 1000 are divisible by 4, this gives 7(x + 4) + 75 = 250. Subtract 75 to get 7(x + 4) = 175, and then expand to obtain 7x + 28 = 175.

Subtracting 28 gives 7x = 147, so x = 21 again. All three methods reach the same solution because they preserve the original equality. Choose a route that simplifies the numbers or the grouped expression clearly.

Note: Dividing a side containing a sum means dividing the whole sum. In the division-by-4 method, both 28(x + 4) and 300 are divided by 4; leaving 300 unchanged would alter the equation.

How do we turn a party budget into an equation?

A word problem describes numerical relationships in sentences. Start by deciding exactly what the unknown represents. Madhubanti’s snacks cost ₹25 per plate, where ₹ denotes rupees. Delivery costs a fixed ₹50, meaning this charge does not change with the number of plates in the problem.

She has ₹500 to spend, and her family contains five members, including herself. Each family member and each invited friend receives one plate. The number of people attending and the number of friends invited are therefore different quantities.

What if the unknown counts all the people?

Worked example 8. Find how many friends Madhubanti can invite with ₹500, snacks at ₹25 per person, a ₹50 delivery charge, and five family members receiving snacks.

Answer: Let p be the total number of people. The equation is 25p + 50 = 500. Subtracting 50 gives 25p = 450, so p = 18. Removing the five family members from this total gives 18 − 5 = 13 friends.

The check includes both parts of the cost: 25 × 18 + 50 = 500. Reporting 18 friends would confuse the total attendance with the invited friends. The definition of p tells us what the solved number means.

What if the unknown counts only the friends?

Let f represent the number of friends. Then f + 5 is the total number receiving snacks. After reserving ₹50 for delivery, ₹450 remains for plates. The equation is 25(f + 5) = 450. Divide by 25 to get f + 5 = 18, then subtract 5 to get f = 13.

The two equations answer the same question using different unknowns. One includes delivery in the equation itself; the other first subtracts delivery from the budget. Both must account for all five family members and the fixed charge.

1. Write what the chosen letter-number represents, including the quantity being counted.
2. Express the cost or total using that unknown and the given information.
3. Set the expression equal to the stated amount and solve systematically.
4. Interpret the result in the situation and check it against all the original data.

How can equations compare savings and divide a total?

Some word problems compare two changing quantities. Others give a total together with a relationship between its parts. In both cases, the equation comes from translating each statement carefully, with a consistent meaning for the unknown.

When will Jahnavi and Sunita have equal savings?

Jahnavi starts with ₹4000 and saves ₹650 each month. Sunita starts with ₹5050 and saves ₹500 each month. Let m represent the number of months after which their savings are equal. Their savings then are 4000 + 650m and 5050 + 500m rupees.

Worked example 9. Find when the two savings amounts become equal using the initial amounts and monthly savings above.

Answer: Form 4000 + 650m = 5050 + 500m. Subtract 500m to get 4000 + 150m = 5050. Subtract 4000 to get 150m = 1050. Divide by 150 to obtain m = 7. Thus their savings become equal after seven months.

A check of both amounts gives 4000 + 650 × 7 = 8550 and 5050 + 500 × 7 = 8550. Sunita’s larger initial amount is matched after Jahnavi’s larger monthly additions have accumulated for the required time.

How can two unknown quantities be expressed using one letter?

Ramesh and Suresh have 60 marbles altogether. Ramesh has 30 more than Suresh. Let y be Suresh’s number of marbles. Ramesh’s number is then y + 30, so a second independent letter is unnecessary for solving this problem.

Worked example 10. Find each boy’s marbles when the total is 60 and Ramesh has 30 more than Suresh.

Answer: Write y + (y + 30) = 60. This simplifies to 2y + 30 = 60. Subtracting 30 and dividing by 2 gives y = 15. Suresh has 15 marbles and Ramesh has 45. Their total is 60 and their difference is 30.

Both conditions must be satisfied: the total and the difference. Finding two numbers with the correct sum alone is insufficient. Expressing one quantity in terms of the other builds the difference into the equation before solving it.

How can equations explain number tricks and generate more equations?

A number trick can hide its starting number while revealing its final result. Algebra records every operation using a letter-number. In Riyaz’s trick, let x be Akash’s starting number. The order is subtract 3, multiply the result by 4, then add 8.

How do we keep track of each step?

Worked example 11. Akash’s final result is 24 after subtracting 3 from his starting number, multiplying by 4 and adding 8. Find his starting number.

Answer: With x as the starting number, 4x − 4 = 24. Write this as 4(x − 1) = 24. Dividing by 4 gives x − 1 = 6, so x = 7. Checking the steps gives 7 − 3 = 4, then 4 × 4 = 16, and finally 16 + 8 = 24.

The brackets matter because the whole result of subtraction is multiplied. Writing 4x − 3 would describe multiplying the starting number first and then subtracting 3, which is a different sequence of operations.

How can we make equations with a known solution?

Let y be an unknown whose intended value is 5. Two equations with this solution are y + 1 = 6 and 3y = 15. Starting from 3y = 15, add 6 to both sides to get 3y + 6 = 21.

Divide both sides of that equation by 3 to obtain y + 2 = 7, then subtract 2 to return to y = 5. The value 5 satisfies each equation in this chain. The reverse chain uses the inverse operations in the opposite order.

Forming equations and solving them are connected activities. Starting with a solution and applying suitable operations produces equations; undoing those operations uncovers the solution. This connection helps explain why a correctly formed chain retains the same answer.

How can we recognise and repair mistakes in working?

A wrong answer often begins with a single operation applied incorrectly. Check the transition between consecutive lines, not just the final value. Ask whether the same operation has been applied to each complete side and whether signs, factors and brackets have been handled consistently.

What goes wrong when terms and factors are confused?

In 4x + 6 = 10, where x is unknown, removing +6 requires 4x = 10 − 6, not 10 + 6. Therefore, 4x = 4 and x = 1. The inverse of adding 6 is subtracting 6.

In 2v − 4 = 6, where v is unknown, 2 is a multiplying factor of v. Subtracting 2 cannot turn 2v into v. Add 4 first to obtain 2v = 10, then divide by 2 to get v = 5.

What goes wrong when only part of a side is divided?

For 3x + 1 = −12, dividing the whole left side by 3 gives x + 1/3, not x + 1. An easier route subtracts 1 first: 3x = −13. Division by 3 then gives x = −13/3.

In 4(4q + 2) = 50, where q is unknown, the outer factor 4 multiplies both terms inside the bracket. Opening the brackets gives 16q + 8 = 50. Subtracting 8 gives 16q = 42, so q = 21/8.

How can checking expose the mistaken answer?

For 7 − 8z = 5, where z is unknown, the line 8z = 2 is valid, but its solution is z = 2/8 = 1/4, not 4. Substituting 1/4 gives 7 − 2 = 5, confirming equality.

These corrections rely on the same properties used in ordinary solving. A term being added is treated differently from a factor being multiplied. A whole sum is treated differently from one of its terms. Substitution provides a direct final test after each repair.

What does the history of algebra show about finding unknowns?

Bījagaṇita, now also known as algebra, uses letter symbols to solve mathematical problems. The word bīja means seed. The comparison is that an answer lies hidden in an unknown number, just as a tree lies hidden in a seed.

How did algebraic ideas and notation develop?

In Chapter 18 of Brāhmasphuṭasiddhānta, dated 628 CE (Common Era), Brahmagupta explained operations on unknown numbers using letters. This chapter was one of the earliest known works in algebra in history.

Indian mathematical ideas were translated into Arabic in the 8th century. They influenced Al-Khwarizmi, who lived in present-day Iraq. Around 825 CE, he wrote Hisab al-jabr wal-muqabala, meaning calculation by restoring and balancing. By the 12th century, his book had been translated into Latin and brought to Europe.

The word al-jabr gave rise to the word algebra. Ancient Indian notation used yā, short for yāvat-tāvat, meaning as much as needed, for an unknown. Known quantities were denoted by rū, from rūpa. A dot above a number indicated a negative number.

How does the horse-price problem become an equation?

Worked example 12. One man has ₹300 and six horses. Another has ten horses and a debt of ₹100. They are equally rich, and each horse has the same price. Find that price.

Answer: Let x be one horse’s price in rupees. Then 300 + 6x = 10x − 100. Add 100 to obtain 400 + 6x = 10x. Subtract 6x to get 400 = 4x. Hence x = 100, so each horse costs ₹100.

This problem appears in Bījgaṇita by Bhāskarāchārya, dated 1150 CE. The debt reduces the second man’s wealth, so it is represented by subtraction. The equality compares their total wealth, including the value of their horses.

How does Brahmagupta’s general formula work?

In Ax + B = Cx + D, x is the unknown; A and C are its known multiplying numbers, and B and D are known added numbers. Brahmagupta’s formula is x = (D − B)/(A − C). It applies when A − C is not zero.

Subtract Cx and B from both sides to get (A − C)x = D − B, then divide by A − C. For the savings problem, this gives m = (5050 − 4000)/(650 − 500) = 7. Algebra generalises the steps used in individual problems.

Glossary

• Unknown quantity — A number or amount whose value is to be found from the information in a problem.
• Letter-number — A letter used to represent a number when writing an expression or equation.
• Algebraic expression — A combination of numbers and letter-numbers connected through arithmetic operations.
• Equation — A mathematical statement expressing equality between two algebraic expressions, written using an equals sign.
• Left Hand Side — The expression written to the left of the equals sign in an equation.
• Right Hand Side — The expression written to the right of the equals sign in an equation.
• Solution — A value of the unknown that makes the two sides of an equation equal.
• Substitution — Replacing a letter-number with a particular number to evaluate an expression or check an equation.
• Trial and error — Trying different values for an unknown and testing which value makes the equation true.
• Inverse operations — Operations that undo one another, such as addition and subtraction, or multiplication and division.
• Additive inverse — The number that combines with a given number by addition to make zero.
• Factor — A number or expression multiplied by another to form a product.
• Divisor — The number by which another number or expression is divided.
• Bījagaṇita — The branch of mathematics, also called algebra, using letter symbols to solve mathematical problems.

Common errors and misconceptions

• Misconception: Removing +6 from 4x + 6 = 10 gives 4x = 16. Correct: Subtract 6 from both sides, obtaining 4x = 4 and x = 1.
• Misconception: Subtracting 2 changes 2v into v. Correct: The 2 is a factor, so divide by 2 when removing it; subtracting 2 is a different operation.
• Misconception: Dividing 3x + 1 by 3 gives x + 1. Correct: Divide every term in the sum; the result is x + 1/3.
• Misconception: Only 4q is multiplied by the outer 4 in 4(4q + 2). Correct: The 2 is also multiplied, giving 16q + 8.
• Misconception: The equation 8z = 2 gives z = 4. Correct: Divide 2 by 8, giving z = 1/4, and check the result by substitution.
• Misconception: Every solution must be a positive whole number. Correct: The solution of 5x − 4 = 7 is 11/5, while 5s = 3s has solution zero.
• Misconception: The party calculation gives 18 invited friends. Correct: Eighteen is the total number receiving snacks, including five family members; the number of friends is 13.

Exam-style questions with model answers

Q1. What is an equation, and what does it mean to solve an equation? [2 marks]

1. An equation states that two algebraic expressions have equal values, using the equals sign between them.
2. Solving it means finding the unknown value or values that make the Left Hand Side equal to the Right Hand Side.

Q2. Solve 5x − 4 = 7, where x is unknown, and check the solution. [3 marks]

1. Add 4 to both sides to remove the subtracted 4. The equation becomes 5x = 7 + 4 = 11.
2. Divide both sides by the non-zero factor 5. This gives x = 11/5, which is the value to test.
3. Substitute 11/5 into the original left side: 5 × (11/5) − 4 = 11 − 4 = 7. It equals the right side, confirming the solution.

Q3. Solve 6y + 7 = 4y + 21, where y is unknown, and verify your answer. [4 marks]

1. Subtract 4y from both sides to collect the terms containing the same unknown: 6y − 4y + 7 = 21, giving 2y + 7 = 21.
2. Subtract 7 from both sides, leaving 2y = 21 − 7 = 14.
3. Divide by 2 to obtain the unknown’s value, y = 7.
4. Check the original equation: its left side is 6 × 7 + 7 = 49, and its right side is 4 × 7 + 21 = 49. The sides agree.

Q4. A matchstick sequence uses 2n + 1 sticks at position n, where n is a positive whole number. Find the position using 99 sticks and explain whether exactly 200 sticks can form an arrangement. [3 marks]

1. For 99 sticks, form 2n + 1 = 99. Subtracting 1 and dividing by 2 gives n = 49.
2. Check the position by calculating 2 × 49 + 1 = 99, so position 49 has the required number of sticks.
3. For 200 sticks, 2n + 1 = 200 gives n = 199/2. This is not a whole-number position, so exactly 200 sticks cannot form an arrangement in this sequence.

Q5. Madhubanti has ₹500 for snacks and delivery. Snacks cost ₹25 per plate and delivery costs a fixed ₹50. Her family has five members, including herself. Each family member and invited friend gets one plate. Form an equation, find the number of friends she can invite, and check the cost. [5 marks]

1. Let p represent the total number of people receiving snacks, including family and friends. The snack cost is therefore 25p rupees.
2. Add the fixed delivery charge of ₹50. Using the full ₹500 budget gives the equation 25p + 50 = 500.
3. Subtract 50 from both sides to get 25p = 450. Divide by 25 to obtain p = 18 people in total.
4. Five of these people are family members. Therefore, the number of friends who can be invited is 18 − 5 = 13.
5. Check using all 18 plates: 25 × 18 + 50 = 450 + 50 = ₹500. This includes every person’s snack and the delivery charge.

Q6. Jahnavi has ₹4000 and saves ₹650 per month. Sunita has ₹5050 and saves ₹500 per month. After how many months will their savings be equal? Form an equation and check both amounts. [5 marks]

1. Let m be the number of months. Jahnavi’s savings after that time are 4000 + 650m rupees, and Sunita’s are 5050 + 500m rupees.
2. For equal savings, write 4000 + 650m = 5050 + 500m. The same number of months is used for both friends.
3. Subtract 500m from both sides, giving 4000 + 150m = 5050. Then subtract 4000 to obtain 150m = 1050.
4. Divide both sides by 150. This gives m = 7, so the required time is seven months.
5. Check both amounts after seven months: Jahnavi has 4000 + 650 × 7 = ₹8550, while Sunita has 5050 + 500 × 7 = ₹8550. Their savings are equal.

Q7. Ramesh and Suresh together have 60 marbles. Ramesh has 30 more marbles than Suresh. Use one unknown to find each boy’s marbles and check both conditions. [4 marks]

1. Let y be Suresh’s number of marbles. Ramesh then has y + 30 marbles, since he has 30 more than Suresh.
2. Use their total to form y + (y + 30) = 60, which simplifies to 2y + 30 = 60.
3. Subtract 30 and divide by 2: 2y = 30, so y = 15. Suresh has 15 marbles and Ramesh has 45.
4. Check both conditions: 15 + 45 = 60, and 45 − 15 = 30. The total and the difference are correct.

Q8. A student solves 4x + 6 = 10 by writing 4x = 10 + 6 and then x = 4. Identify the mistake and give the correct solution, where x is unknown. [2 marks]

1. The added 6 must be removed by subtracting 6 from both sides, not by adding it to 10.
2. The correct working is 4x = 10 − 6 = 4, so x = 1. Checking gives 4 × 1 + 6 = 10.

Key takeaways

• An equation states equality between two expressions; a solution makes the Left Hand Side equal to the Right Hand Side.
• Define the unknown before forming an equation, so the final number can be interpreted correctly in its situation.
• Equations can often be solved by performing the same operation on both sides until the unknown’s value becomes evident.
• Addition and subtraction undo each other; multiplication and division undo each other when the division is permitted.
• When an unknown appears on both sides, collecting its terms on one side can simplify the equation.
• Brackets group an expression, so an outside multiplying factor applies to every term inside the bracket.
• Check solutions in the original equation, and check all conditions when the equation comes from a word problem.
• Algebra connects numerical and shape patterns, everyday problems, and general procedures for finding unknown values.

Test yourself

What does 2e mean when e is the weight of one egg?

It means twice the weight of one egg, representing the combined weight of two equally heavy eggs.

Why can trial and error be inefficient?

It can require several guesses and checks before finding a value that makes the two sides equal.

If u/15 = 6, where u is unknown, how do you find u?

Multiply both sides by 15 to undo division by 15. This gives u = 6 × 15 = 90.

Why does 5s = 3s have the solution s = 0?

Subtracting 3s gives 2s = 0. Dividing by 2 gives s = 0, which makes both original sides zero.

Why does x + 4 = x + 5 have no solution?

Subtracting x from both sides would give 4 = 5, which is false for every possible value of x.

What is the first useful step in solving 28(x + 4) + 300 = 1000?

Subtract 300 from both sides to get 28(x + 4) = 700, then divide by 28.

In Riyaz’s trick, subtract 3, multiply by 4 and add 8. If the final answer is 24, what was the starting number?

The starting number was 7: subtracting 3 gives 4, multiplying by 4 gives 16, and adding 8 gives 24.

In Ax + B = Cx + D, what do the letters mean, and when can Brahmagupta’s formula be used?

The unknown is x; A and C multiply it, while B and D are known added numbers. Use x = (D − B)/(A − C) when A − C is non-zero.