Namaste, my dear students! Welcome to today's mathematics lesson. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 7, which is called "Finding the Unknown." Now, isn't that interesting? We are going to learn how to find values that we don't know yet, using the powerful tools of algebra. Are you ready? Let's begin!
Imagine you have a weighing scale in front of you. On one side, there are some weights, and on the other side, there is a sack whose weight you don't know. The scale is balanced, which means both sides have equal weight. Your task is to find the weight of that unknown sack. This is exactly what we are going to learn in the first part of this chapter. We call these "unknown weights" problems.
Look at the figures in your textbook. In Figure 7.1, you see a simple balance with some weights on both sides. The numbers written there represent the same units of weight. Your job is to find the unknown weight in each case. Now, think about this carefully. If the scale is balanced, then the total weight on the left side must equal the total weight on the right side. This is exactly like an equation! When we don't know a weight, we can represent it with a letter, like x or y or any symbol. This letter represents an unknown value that we need to find.
Let me tell you an important idea. If we remove equal weights from both plates of a balanced scale, the scale still remains balanced. This is a fundamental property that will help us solve many problems. Similarly, if we remove objects so that all the unknown weights are on only one plate, the problem becomes much easier to solve. This is like simplifying an equation by moving terms from one side to the other.
Now, let's look at another interesting scenario. Consider a sequence of matchstick arrangements. You might have seen this pattern in earlier classes. The first arrangement has 2 matchsticks in each row plus 1 in the middle, making 3 matchsticks in total. The second arrangement has 2 times 2 plus 1, which equals 5 matchsticks. The third arrangement has 2 times 3 plus 1, which equals 7 matchsticks. Do you see the pattern? The number of matchsticks in the nth position is given by the expression 2n + 1.
Now, here comes an interesting question. Jasmine wants to make a matchstick arrangement using exactly 99 sticks. What will be the position number of this arrangement in the sequence? To answer this, we need to find the value of n such that 2n + 1 equals 99. So we need to solve the equation 2n + 1 = 99. Can you find the value of n? Let's think. If n is 49, then 2 times 49 plus 1 equals 98 plus 1, which is 99. So n equals 49. That means the 49th arrangement in the sequence will have exactly 99 matchsticks.
Now, here's another question. Is it possible to make a matchstick arrangement that appears in this sequence using exactly 200 sticks? Let's check. If 2n + 1 = 200, then 2n = 199, which gives n = 199/2 = 99.5. But n must be a whole number because it represents a position in the sequence. Since 99.5 is not a whole number, it is not possible to make such an arrangement using exactly 200 sticks. This is an important insight - sometimes, the equation might have a solution that is not a whole number, and in such cases, the real-life situation might not be possible.
Now, let's talk about something very important in mathematics. A statement of equality between two algebraic expressions is called an equation. In modern mathematics, we write an equation as two algebraic expressions with an equal sign '=' between them. For example, 3x + 4 = 7 is an equation. Here, 3x + 4 is the left-hand side, and 7 is the right-hand side. The letter x is called an unknown or a variable. Our goal is to find the value of x such that the left-hand side equals the right-hand side.
The process of finding the value or values of the letter-numbers for which the equality holds, or for which the value of the left-hand side becomes equal to the right-hand side, is called solving the equation. As we saw in the matchstick problem, framing an equation using an unknown quantity as a letter-number helps us find its value.
Now, let's practice framing equations for the weighing scale problems. In Figure 7.6, let's denote the weight of one fried egg as e. Since each slice of bread weighs 2 units, we have three slices of bread weighing 2 + 2 + 2 = 6 on one side, and two fried eggs weighing e + e on the other side. Since they are equal, we have 6 = e + e, or 2e = 6. This is our equation! Similarly, for Figure 7.7, we can denote the weight of one item as y and form the equation 4 + 2y = 16. Once we have these equations, we can solve them to find the unknown weights.
Now, let's discuss how to solve equations systematically. One way is the trial and error method, where we substitute different values in place of the letter-number and check which value makes the left-hand side equal to the right-hand side. For example, consider the equation 2n + 1 = 99. If we substitute n = 5, we get 2 × 5 + 1 = 11, which is far from 99. If we try n = 10, we get 21. Still not 99. Let's try n = 30. We get 61. Still lower. n = 40 gives us 81. Getting closer! When n = 50, we get 101, which is a bit too high. Finally, when n = 49, we get 99! So the solution is n = 49. This method works, but it can be inefficient, especially when the numbers are large.
So, is there a better way? Yes, there is! Remember the weighing scale? When we remove equal weights from both plates, the scale remains balanced. Similarly, for equations, if we perform the same operation on both sides, the equality is preserved. This is a fundamental property of equations. Let's understand this with some examples.
Consider the equation 15 + 8 = 23. If we add, subtract, multiply, or divide the same number on both sides, will it still preserve the equality? Yes, it will! For example, if we add 10 to both sides, we get 15 + 8 + 10 = 23 + 10, which is 33 = 33. The equality holds. This is because the left-hand side and the right-hand side of an equation have the same value, so performing the same operation on both sides will clearly not change their equality.
Now, let's use this idea to solve equations. Let's solve the equation 5x – 4 = 7. Our goal is to get x alone on one side. First, we need to remove the – 4 from the left-hand side. We can do this by adding 4 to both sides. So, 5x – 4 + 4 = 7 + 4, which gives us 5x = 11. Now, we need to remove the factor 5 from the left-hand side. We can do this by dividing both sides by 5. So, 5x/5 = 11/5, which gives us x = 11/5. That's the solution!
Now, let's check if this is correct. We substitute x = 11/5 in the original equation 5x – 4 = 7. The left-hand side becomes 5 × (11/5) – 4 = 11 – 4 = 7, which is exactly equal to the right-hand side. So, our solution is correct!
Let's try another example. Solve the equation 11y + (–5) = 61. Here, we need to remove the –5 from the left-hand side. We can do this by subtracting –5 from both sides, which is the same as adding 5. So, 11y + (–5) – (–5) = 61 – (–5), which gives us 11y = 66. Now, we can see that 11 × 6 = 66, so y = 6. We can also divide both sides by 11 to get y = 66 ÷ 11 = 6. Let's check: 11 × 6 + (–5) = 66 – 5 = 61. Perfect!
Now, let's look at an equation where the unknown appears on both sides. Solve 6y + 7 = 4y + 21. First, we need to bring all the terms with y to one side. Let's subtract 4y from both sides: 6y + 7 – 4y = 4y + 21 – 4y, which gives us 2y + 7 = 21. Now, subtract 7 from both sides: 2y + 7 – 7 = 21 – 7, which gives us 2y = 14. So, y = 7. Let's check: 6 × 7 + 7 = 42 + 7 = 49, and 4 × 7 + 21 = 28 + 21 = 49. Both sides are equal, so our solution is correct!
Now, let's make some important observations about solving equations. When a term that is added or subtracted on one side of an equation is removed from that side, its additive inverse should appear as a term on the other side for the equality to hold. For example, 2y + 7 = 21 becomes 2y = 21 – 7. We subtracted 7 from both sides, which is the same as adding the additive inverse of 7, which is –7.
Similarly, if one side of an equation is the product of two or more numbers or expressions, and we remove one of the factors, then the other side should be divided by this factor for the equality to hold. For example, 2y = 14 becomes y = 14 ÷ 2.
And if one side of an equation is the quotient of two numbers or expressions, and we remove the divisor, then the other side should be multiplied by this divisor for the equality to hold. For example, u/15 = 6 becomes u = 6 × 15.
These are powerful techniques that will help you solve many equations quickly and efficiently. Remember, the key idea is to perform the same operation on both sides to isolate the unknown.
Now, let's apply what we've learned to solve some real-life problems. This is where algebra becomes really useful!
Example 7: Ranjana creates a sequence of arrangements with square tiles. She has step 1 with 4 tiles, step 2 with 7 tiles, step 3 with 10 tiles, and step 4 with 13 tiles. Can she extend the sequence and make an arrangement using 100 tiles? If yes, which step in the sequence will it be?
Let's look at the pattern. In step 1, we have 1 + 1 + 1 + 1 = 4, which is 3 × 1 + 1 = 4. In step 2, we have 2 + 2 + 2 + 1 = 7, which is 3 × 2 + 1 = 7. In step 3, we have 3 + 3 + 3 + 1 = 10, which is 3 × 3 + 1 = 10. So, in step k, we would have 3k + 1 tiles. To find if an arrangement with 100 tiles is possible, we need to solve the equation 3k + 1 = 100. Subtracting 1 from both sides gives 3k = 99, so k = 33. Yes, it is possible! The 33rd step will have exactly 100 tiles.
Example 8: Madhubanti wants to organize a party. Each plate of snacks costs ₹25, and there is an additional fixed delivery charge of ₹50. She has ₹500 to spend. There are 5 members in her family, including herself. How many friends can she invite?
Let's solve this using algebra. Let p be the total number of people who can attend the party, including Madhubanti and her family. The total cost would be 25p + 50. Since this should equal 500, we have the equation 25p + 50 = 500. Subtracting 50 from both sides gives 25p = 450. Dividing by 25 gives p = 18. So, 18 people can attend the party. Since there are 5 family members, she can invite 18 – 5 = 13 friends.
Alternatively, let f be the number of friends she can invite. Then the total number of people is f + 5, and the cost is 25(f + 5) = 450 (since ₹50 is for delivery). Dividing by 25 gives f + 5 = 18, so f = 13. Same answer!
Example 9: Two friends, Jahnavi and Sunita, want to save money. Jahnavi starts with ₹4000 and saves ₹650 per month. Sunita starts with ₹5050 and saves ₹500 per month. After how many months will they have the same amount of money?
Let m be the number of months after which their savings are equal. Jahnavi's savings after m months will be 4000 + 650m. Sunita's savings after m months will be 5050 + 500m. Since these are equal, we have 4000 + 650m = 5050 + 500m. Subtracting 500m from both sides gives 4000 + 150m = 5050. Subtracting 4000 from both sides gives 150m = 1050. Dividing by 150 gives m = 7. So, after 7 months, they will have the same amount of money. Let's verify: Jahnavi will have 4000 + 650 × 7 = 4000 + 4550 = 8550. Sunita will have 5050 + 500 × 7 = 5050 + 3500 = 8550. Perfect!
Example 10: Solve the equation 28(x + 4) + 300 = 1000. There are multiple ways to solve this. Let's look at one method. First, subtract 300 from both sides: 28(x + 4) = 700. Now, divide both sides by 28: x + 4 = 700 ÷ 28 = 25. So, x = 25 – 4 = 21. That's the solution!
Example 11: Riyaz created a math trick. He asked Akash to think of a number, subtract 3, multiply by 4, add 8, and reveal the final answer, which was 24. What was the starting number?
Let the starting number be x. After subtracting 3, we get x – 3. After multiplying by 4, we get 4(x – 3) = 4x – 12. After adding 8, we get 4x – 12 + 8 = 4x – 4. Since the final answer is 24, we have 4x – 4 = 24. Adding 4 to both sides gives 4x = 28. Dividing by 4 gives x = 7. So, Akash thought of the number 7. Notice that the final answer minus 4, divided by 4, plus 1 gives the starting number. There's a pattern here that you can use to create your own math tricks!
Example 12: Ramesh and Suresh have 60 marbles between them. Ramesh has 30 more marbles than Suresh. How many marbles does each boy have?
Let the number of marbles with Suresh be y. Then Ramesh has y + 30 marbles. The total is y + (y + 30) = 60, which gives 2y + 30 = 60. Subtracting 30 from both sides gives 2y = 30, so y = 15. That means Suresh has 15 marbles, and Ramesh has 15 + 30 = 45 marbles. Let's check: 15 + 45 = 60, and 45 is indeed 30 more than 15. Perfect!
Now, let's talk about generating equations. Can we do the reverse? Can we write equations using a given value of the letter-number? For example, write equations whose solution is y = 5. There are many possibilities! y + 1 = 6 is one equation. 3y = 15 is another. We can create many equations that have the same solution. This is a great way to practice forming equations.
Now, let's look at the section called "Mind the Mistake, Mend the Mistake." This is very important because we often make mistakes when solving equations. Let's go through some examples together.
In the first problem, we have 4x + 6 = 10. The solution given is: 4x = 10 + 6, so 4x = 16, and x = 4. Is this correct? Let's check. If we add 6 to both sides, we get 4x = 16. That's wrong! We should subtract 6 from both sides, not add. The correct step is 4x = 10 – 6 = 4, so x = 1. Let's verify: 4 × 1 + 6 = 4 + 6 = 10. Yes, that's correct!
In the second problem, we have 7 – 8z = 5. The solution given is: 8z = 7 – 5, so 8z = 2, and z = 4. Is this correct? Let's check. If we rearrange, we get –8z = 5 – 7 = –2, so 8z = 2, and z = 2/8 = 1/4. The given solution z = 4 is wrong! The correct solution is z = 1/4.
In the third problem, we have 2v – 4 = 6. The solution given is: v – 4 = 6 – 2, so v – 4 = 4, and v = 8. Is this correct? Wait, they divided only part of the equation by 2! That's not how we solve equations. We need to divide the entire left-hand side and the entire right-hand side by 2. So, 2v – 4 = 6 becomes v – 2 = 3 after dividing by 2. Then v = 5. Let's verify: 2 × 5 – 4 = 10 – 4 = 6. Yes, that's correct!
This shows how important it is to be careful when solving equations. One small mistake can give us the wrong answer. Always check your solution by substituting it back into the original equation!
Now, let's take a moment to appreciate the rich history of algebra. Did you know that algebra was developed in ancient India? The area of mathematics dealing with unknown numbers was called "bījagaṇita," which is now known as algebra. The word "bīja" means seed. Just as a tree is hidden inside a seed, the answer to a problem is hidden inside an unknown number. Solving the problem is like helping the tree grow—step by step, we discover what is hidden.
Brahmagupta, a famous mathematician from ancient India, explained how to add, subtract, and multiply unknown numbers using letters, similar to how we use x or y today. This was one of the earliest known works in algebra in history. In the 8th century, Indian mathematical ideas were translated into Arabic and influenced a well-known mathematician named Al-Khwarizmi, who lived in present-day Iraq. He wrote a book called "Hisab al-jabr wal-muqabala," which means "calculation by restoring and balancing." The word "al-jabr" gave us the word "algebra" that we use today.
Ancient Indian mathematicians used distinct symbols like yā, kā, nī, pī, lo, etc., for different unknowns. The symbol yā was short for "yāvat-tāvat," meaning "as much as needed." Others like kā and nī referred to the first letters of the names of colours—kālaka (black), nīlaka (blue), and so on. The known quantities were called rūpa and were denoted by rū.
For example, in ancient Indian notation, 2x + 1 would be written as "yā 2 rū 1," where yā indicates the unknown and rū indicates the known quantity. If there was a dot over a number, it indicated that it was negative. Isn't that fascinating?
Now, let's summarize what we have learned in this chapter.
First, we learned that an algebraic equation is a mathematical statement that indicates the equality of two algebraic expressions. We use letters like x, y, z to represent unknown values.
Second, we learned that when the same operation is performed on both sides of an equation, equality is maintained. This is the key to solving equations.
Third, we learned that finding a solution to an equation means finding the values of the unknowns such that the left-hand side equals the right-hand side.
Fourth, we learned different methods to solve equations. We can use trial and error, which works but can be slow. We can also use systematic methods, where we perform the same operation on both sides to isolate the unknown.
Fifth, we learned about transposing terms. When a term is added on one side, we can subtract it from both sides to remove it. When a term is subtracted, we can add it to both sides. When a term is multiplied, we can divide both sides by it. When a term is divided, we can multiply both sides by it.
Sixth, we learned how to solve equations with unknowns on both sides. We bring all the unknown terms to one side and all the known terms to the other side.
Seventh, we learned how to check our solutions by substituting the value back into the original equation.
Eighth, we learned how to apply algebra to solve real-life problems, like finding the number of friends we can invite to a party or when two friends will have the same amount of money.
And finally, we learned about the rich history of algebra in ancient India and how it spread to other parts of the world.
Algebra is a powerful tool that helps us solve problems in our daily lives. It allows us to generalize patterns and express them using precise mathematical language. By studying algebra, we learn to think logically and systematically. I hope you enjoyed learning this chapter as much as I enjoyed teaching it!
Now, before I say goodbye, let me leave you with a fun puzzle. Think of any number. Multiply it by 2. Add 10. Divide by 2. Now subtract the original number you thought of. Finally, add 3. I predict that you now have 8. Am I correct? Can you explain why this trick works? Try to create your own such tricks using algebra!
Thank you for being such wonderful students. Keep practicing, keep exploring, and keep loving mathematics. See you next time!