Welcome dear students! Today we are going to learn about Simple Equations from Class 7 Maths. The teacher has said that she would be starting a new chapter in mathematics and it is going to be simple equations. Appu, Sarita and Ameena have revised what they learnt in algebra chapter in Class 6. Have you? Appu, Sarita and Ameena are excited because they have constructed a game which they call mind reader and they want to present it to the whole class. The teacher appreciates their enthusiasm and invites them to present their game. Ameena begins; she asks Sara to think of a number, multiply it by 4 and add 5 to the product. Then, she asks Sara to tell the result. She says it is 65. Ameena instantly declares that the number Sara had thought of is 15. Sara nods. The whole class including Sara is surprised. It is Appu’s turn now. He asks Balu to think of a number, multiply it by 10 and subtract 20 from the product. He then asks Balu what his result is? Balu says it is 50. Appu immediately tells the number thought by Balu. It is 7, Balu confirms it. Everybody wants to know how the mind reader presented by Appu, Sarita and Ameena works. Can you see how it works? After studying this chapter and chapter 10, you will very well know how the game works.
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Let us take Ameena’s example. Ameena asks Sara to think of a number. Ameena does not know the number. For her, it could be anything 1, 2, 3, up to 11, up to 100, and so on. Let us denote this unknown number by a letter, say x. You may use y or t or some other letter in place of x. It does not matter which letter we use to denote the unknown number Sara has thought of. When Sara multiplies the number by 4, she gets 4x. She then adds 5 to the product, which gives 4x + 5. The value of 4x + 5 depends on the value of x. Thus if x = 1, 4x + 5 = 4 × 1 + 5 = 9. This means that if Sara had 1 in her mind, her result would have been 9. Similarly, if she thought of 5, then for x = 5, 4x + 5 = 4 × 5 + 5 = 25. Thus if Sara had chosen 5, the result would have been 25.
To find the number thought by Sara let us work backward from her answer 65. We have to find x such that 4x + 5 = 65. We will refer to this as equation 4.1. Solution to the equation will give us the number which Sara held in her mind. Let us similarly look at Appu’s example. Let us call the number Balu chose as y. Appu asks Balu to multiply the number by 10 and subtract 20 from the product. That is, from y, Balu first gets 10y and from there 10y – 20. The result is known to be 50. Therefore, 10y – 20 = 50. We will call this equation 4.2. The solution of this equation will give us the number Balu had thought of.
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Note, equation 4.1 and equation 4.2 are equations. Let us recall what we learnt about equations in Class 6. An equation is a condition on a variable. In equation 4.1, the variable is x; in equation 4.2, the variable is y. The word variable means something that can vary, that is, change. A variable takes on different numerical values; its value is not fixed. Variables are denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, and so on. From variables, we form expressions. The expressions are formed by performing operations like addition, subtraction, multiplication and division on the variables. From x, we formed the expression 4x + 5. For this, first we multiplied x by 4 and then added 5 to the product. Similarly, from y, we formed the expression 10y – 20. For this, we multiplied y by 10 and then subtracted 20 from the product. All these are examples of expressions. The value of an expression thus formed depends upon the chosen value of the variable. As we have already seen, when x = 1, 4x + 5 = 9; when x = 5, 4x + 5 = 25. Similarly, when x = 15, 4x + 5 = 4 × 15 + 5 = 65; when x = 0, 4x + 5 = 4 × 0 + 5 = 5.
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Equation 4.1 is a condition on the variable x. It states that the value of the expression 4x + 5 is 65. The condition is satisfied when x = 15. It is the solution to the equation 4x + 5 = 65. When x = 5, 4x + 5 = 25 and not 65. Thus x = 5 is not a solution to the equation. Similarly, x = 0 is not a solution to the equation. No value of x other than 15 satisfies the condition 4x + 5 = 65. The value of the expression 10y – 20 depends on the value of y.
Try These: Verify this by giving five different values to y and finding for each y the value of 10y – 20. From the different values you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition is met. For instance, if y = 5, 10y – 20 = 30. If y = 6, 10y – 20 = 40. If y = 7, 10y – 20 = 50. So y = 7 is the solution.
In an equation there is always an equality sign. The equality sign shows that the value of the expression to the left of the sign, the left hand side or LHS, is equal to the value of the expression to the right of the sign, the right hand side or RHS. In equation 4.1, the LHS is 4x + 5 and the RHS is 65. In equation 4.2, the LHS is 10y – 20 and the RHS is 50. If there is some sign other than the equality sign between the LHS and the RHS, it is not an equation. Thus, 4x + 5 > 65 is not an equation. It says that the value of 4x + 5 is greater than 65. Similarly, 4x + 5 < 65 is not an equation. It says that the value of 4x + 5 is smaller than 65.
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In equations, we often find that the RHS is just a number. In equation 4.1, it is 65 and in equation 4.2, it is 50. But this need not be always so. The RHS of an equation may be an expression containing the variable. For example, the equation 4x + 5 = 6x – 25 has the expression 4x + 5 on the left and 6x – 25 on the right of the equality sign. In short, an equation is a condition on a variable. The condition is that two expressions should have equal value. Note that at least one of the two expressions must contain the variable. We also note a simple and useful property of equations. The equation 4x + 5 = 65 is the same as 65 = 4x + 5. Similarly, the equation 6x – 25 = 4x + 5 is the same as 4x + 5 = 6x – 25. An equation remains the same, when the expressions on the left and on the right are interchanged. This property is often useful in solving equations.
Let us look at Example 1. Write the following statements in the form of equations. (i) The sum of three times x and 11 is 32. (ii) If you subtract 5 from 6 times a number, you get 7. (iii) One fourth of m is 3 more than 7. (iv) One third of a number plus 5 is 8. Solution for (i): Three times x is 3x. Sum of 3x and 11 is 3x + 11. The sum is 32. The equation is 3x + 11 = 32. Solution for (ii): Let us say the number is z; z multiplied by 6 is 6z. Subtracting 5 from 6z, one gets 6z – 5. The result is 7. The equation is 6z – 5 = 7. Solution for (iii): One fourth of m is m/4. It is greater than 7 by 3. This means the difference of m/4 – 7 is 3. The equation is m/4 – 7 = 3. Solution for (iv): Take the number to be n. One third of n is n/3. This one-third plus 5 is n/3 + 5. It is 8. The equation is n/3 + 5 = 8.
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Now, Example 2. Convert the following equations in statement form. (i) x – 5 = 9 (ii) 5p = 20 (iii) 3n + 7 = 1 (iv) m/5 – 2 = 6 Solution for (i): Taking away 5 from x gives 9. Solution for (ii): Five times a number p is 20. Solution for (iii): Add 7 to three times n to get 1. Solution for (iv): You get 6, when you subtract 2 from one-fifth of a number m. What is important to note is that for a given equation, not just one, but many statement forms can be given. For example, for the first equation above, you can say: Subtract 5 from x, you get 9. Or, the number x is 5 more than 9. Or, the number x is greater by 5 than 9. Or, the difference between x and 5 is 9, and so on.
Try These: Write at least one other form for each equation (ii), (iii) and (iv). For (ii) 5p = 20, you could say: Twenty divided by 5 gives p. For (iii) 3n + 7 = 1, you could say: Three times a number n is 6 less than 1. For (iv) m/5 – 2 = 6, you could say: Two subtracted from one-fifth of m gives 6.
Consider the following situation in Example 3. Raju’s father’s age is 5 years more than three times Raju’s age. Raju’s father is 44 years old. Set up an equation to find Raju’s age. Solution: We do not know Raju’s age. Let us take it to be y years. Three times Raju’s age is 3y years. Raju’s father’s age is 5 years more than 3y; that is, Raju’s father is 3y + 5 years old. It is also given that Raju’s father is 44 years old. Therefore, 3y + 5 = 44. We will call this equation 4.3. This is an equation in y. It will give Raju’s age when solved.
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Example 4. A shopkeeper sells mangoes in two types of boxes, one small and one large. A large box contains as many as 8 small boxes plus 4 loose mangoes. Set up an equation which gives the number of mangoes in each small box. The number of mangoes in a large box is given to be 100. Solution: Let a small box contain m mangoes. A large box contains 4 more than 8 times m, that is, 8m + 4 mangoes. But this is given to be 100. Thus 8m + 4 = 100. We will call this equation 4.4. You can get the number of mangoes in a small box by solving this equation.
Now let us move to Exercise 4.1. Question 1 asks to check whether the given value satisfies the equation. I will read each row and tell you whether the equation is satisfied. (i) x + 3 = 0, value x = 3. Substituting 3 gives 3 + 3 = 6, not 0. So, No. (ii) x + 3 = 0, value x = 0. 0 + 3 = 3, not 0. So, No. (iii) x + 3 = 0, value x = –3. –3 + 3 = 0. So, Yes. (iv) x – 7 = 1, value x = 7. 7 – 7 = 0, not 1. So, No. (v) x – 7 = 1, value x = 8. 8 – 7 = 1. So, Yes. (vi) 5x = 25, value x = 0. 5 × 0 = 0, not 25. So, No. (vii) 5x = 25, value x = 5. 5 × 5 = 25. So, Yes. (viii) 5x = 25, value x = –5. 5 × –5 = –25, not 25. So, No. (ix) m/3 = 2, value m = –6. –6 / 3 = –2, not 2. So, No. (x) m/3 = 2, value m = 0. 0 / 3 = 0, not 2. So, No. (xi) m/3 = 2, value m = 6. 6 / 3 = 2. So, Yes.
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Question 2: Check whether the value given in the brackets is a solution to the given equation or not. (a) n + 5 = 19 (n = 1). 1 + 5 = 6, not 19. So, No. (b) 7n + 5 = 19 (n = –2). 7 × –2 is –14. –14 + 5 = –9, not 19. So, No. (c) 7n + 5 = 19 (n = 2). 7 × 2 is 14. 14 + 5 = 19. So, Yes. (d) 4p – 3 = 13 (p = 1). 4 × 1 is 4. 4 – 3 = 1, not 13. So, No. (e) 4p – 3 = 13 (p = –4). 4 × –4 is –16. –16 – 3 = –19, not 13. So, No. (f) 4p – 3 = 13 (p = 0). 4 × 0 is 0. 0 – 3 = –3, not 13. So, No.
Question 3: Solve by trial and error method. (i) 5p + 2 = 17. Try p = 1: 5 + 2 = 7. Try p = 2: 10 + 2 = 12. Try p = 3: 15 + 2 = 17. So p = 3. (ii) 3m – 14 = 4. Try m = 4: 12 – 14 = –2. Try m = 5: 15 – 14 = 1. Try m = 6: 18 – 14 = 4. So m = 6.
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Question 4: Write equations for the following statements. (i) The sum of numbers x and 4 is 9. Equation: x + 4 = 9. (ii) 2 subtracted from y is 8. Equation: y – 2 = 8. (iii) Ten times a is 70. Equation: 10a = 70. (iv) The number b divided by 5 gives 6. Equation: b/5 = 6. (v) Three-fourth of t is 15. Equation: 3t/4 = 15. (vi) Seven times m plus 7 gets you 77. Equation: 7m + 7 = 77. (vii) One-fourth of a number x minus 4 gives 4. Equation: x/4 – 4 = 4. (viii) If you take away 6 from 6 times y, you get 60. Equation: 6y – 6 = 60. (ix) If you add 3 to one-third of z, you get 30. Equation: z/3 + 3 = 30.
Question 5: Write the following equations in statement forms. (i) p + 4 = 15. Statement: Four added to p gives 15. (ii) m – 7 = 3. Statement: Seven subtracted from m gives 3. (iii) 2m = 7. Statement: Two times m gives 7. (iv) m/5 = 3. Statement: One-fifth of m is 3. (v) 3m/5 = 6. Statement: Three-fifths of m is 6. (vi) 3p + 4 = 25. Statement: Three times p plus 4 gives 25. (vii) 4p – 2 = 18. Statement: Four times p minus 2 gives 18. (viii) p/2 + 2 = 8. Statement: Half of p plus 2 gives 8.
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Question 6: Set up an equation in the following cases. (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. Take m to be the number of Parmit’s marbles. Five times m is 5m. Seven more than that is 5m + 7. So, 5m + 7 = 37. (ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. Take Laxmi’s age to be y years. Three times y is 3y. Four years older is 3y + 4. So, 3y + 4 = 49. (iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. Take the lowest score to be l. Twice l is 2l. Plus 7 is 2l + 7. So, 2l + 7 = 87. (iv) In an isosceles triangle, the vertex angle is twice either base angle. Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees. The vertex angle is 2b. The sum of angles is b + b + 2b = 180. So, 4b = 180.
Now let us learn about solving an equation. Consider an equality 8 – 3 = 4 + 1. We will refer to this as equation 4.5. The equality holds, since both its sides are equal, each is equal to 5. Let us now add 2 to both sides; as a result LHS = 8 – 3 + 2 = 5 + 2 = 7. RHS = 4 + 1 + 2 = 5 + 2 = 7. Again the equality holds. Thus if we add the same number to both sides of an equality, it still holds. Let us now subtract 2 from both the sides; as a result, LHS = 8 – 3 – 2 = 5 – 2 = 3. RHS = 4 + 1 – 2 = 5 – 2 = 3. Again, the equality holds. Thus if we subtract the same number from both sides of an equality, it still holds.
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Similarly, if we multiply or divide both sides of the equality by the same non-zero number, it still holds. For example, let us multiply both the sides of the equality by 3, we get LHS = 3 × (8 – 3) = 3 × 5 = 15. RHS = 3 × (4 + 1) = 3 × 5 = 15. The equality holds. Let us now divide both sides of the equality by 2. LHS = (8 – 3) ÷ 2 = 5 ÷ 2 = 5/2. RHS = (4 + 1) ÷ 2 = 5 ÷ 2 = 5/2. Again, the equality holds. If we take any other equality, we shall find the same conclusions. Suppose we do not observe these rules. Specifically, suppose we add different numbers to the two sides of an equality. We shall find in this case that the equality does not hold. For example, let us take again equation 4.5, add 2 to the LHS and 3 to the RHS. The new LHS is 8 – 3 + 2 = 5 + 2 = 7 and the new RHS is 4 + 1 + 3 = 5 + 3 = 8. The equality does not hold, because the new LHS and RHS are not equal. Thus if we fail to do the same mathematical operation with same number on both sides of an equality, the equality may not hold. The equality that involves variables is an equation. These conclusions are also valid for equations, as in each equation variable represents a number only.
Often an equation is said to be like a weighing balance. Doing a mathematical operation on an equation is like adding weights to or removing weights from the pans of a weighing balance. In your textbook, you will see a diagram of a weighing balance. The left pan is clearly labeled L.H.S., and the right pan is labeled R.H.S. Equal weights are placed in both pans, keeping the balance arm perfectly horizontal. This visually represents how both sides of an equation must be equal. If we add the same weights to both the pans, the arm remains horizontal. Similarly, if we remove the same weights from both the pans, the arm remains horizontal. On the other hand if we add different weights to the pans or remove different weights from them, the balance is tilted; that is, the arm of the balance does not remain horizontal. We use this principle for solving an equation.
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Let us take some examples. Consider the equation x + 3 = 8. We will call this equation 4.6. We shall subtract 3 from both sides of this equation. The new LHS is x + 3 – 3 = x and the new RHS is 8 – 3 = 5. Since this does not disturb the balance, we have new LHS = new RHS or x = 5, which is exactly what we want, the solution of the equation. To confirm whether we are right, we shall put x = 5 in the original equation. We get LHS = x + 3 = 5 + 3 = 8, which is equal to the RHS as required. By doing the right mathematical operation on both the sides of the equation, we arrived at the solution.
Let us look at another equation x – 3 = 10. We will call this equation 4.7. What should we do here? We should add 3 to both the sides. By doing so, we shall retain the balance and also the LHS will reduce to just x. New LHS = x – 3 + 3 = x. New RHS = 10 + 3 = 13. Therefore, x = 13, which is the required solution. By putting x = 13 in the original equation we confirm that the solution is correct: LHS of original equation = x – 3 = 13 – 3 = 10. This is equal to the RHS as required.
Similarly, let us look at the equations 5y = 35, which is equation 4.8, and m/2 = 5, which is equation 4.9. In the first case, we shall divide both the sides by 5. This will give us just y on LHS. New LHS = 5y / 5 = y. New RHS = 35 / 5 = 7. Therefore, y = 7. This is the required solution. We can substitute y = 7 in the equation and check that it is satisfied. In the second case, we shall multiply both sides by 2. This will give us just m on the LHS. The new LHS = m/2 × 2 = m. The new RHS = 5 × 2 = 10. Hence, m = 10. It is the required solution. You can check whether the solution is correct. One can see that in the above examples, the operation we need to perform depends on the equation. Our attempt should be to get the variable in the equation separated. Sometimes, for doing so we may have to carry out more than one mathematical operation.
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Let us solve some more equations. Example 5. Solve part (a): 3n + 7 = 25. We will call this equation 4.10. Solution: We go stepwise to separate the variable n on the LHS of the equation. The LHS is 3n + 7. We shall first subtract 7 from it so that we get 3n. From this, in the next step we shall divide by 3 to get n. Remember we must do the same operation on both sides of the equation. Therefore, subtracting 7 from both sides, 3n + 7 – 7 = 25 – 7. This gives 3n = 18. Now divide both sides by 3, 3n / 3 = 18 / 3. This gives n = 6, which is the solution.
Part (b): 2p – 1 = 23. We will call this equation 4.11. Solution: What should we do here? First we shall add 1 to both the sides: 2p – 1 + 1 = 23 + 1. This gives 2p = 24. Now divide both sides by 2, we get 2p / 2 = 24 / 2. This gives p = 12, which is the solution. One good practice you should develop is to check the solution you have obtained. Let us put the solution p = 12 back into the equation. LHS = 2p – 1 = 2 × 12 – 1 = 24 – 1 = 23, which equals RHS. The solution is thus checked for its correctness.
We are now in a position to go back to the mind-reading game presented by Appu, Sarita, and Ameena and understand how they got their answers. For this purpose, let us look at equation 4.1 and equation 4.2. First consider equation 4.1: 4x + 5 = 65. Subtracting 5 from both sides, 4x + 5 – 5 = 65 – 5. That is, 4x = 60. Divide both sides by 4; this will separate x. We get 4x / 4 = 60 / 4 or x = 15, which is the solution. Now consider equation 4.2: 10y – 20 = 50. Adding 20 to both sides, we get 10y – 20 + 20 = 50 + 20 or 10y = 70. Dividing both sides by 10, we get 10y / 10 = 70 / 10 or y = 7, which is the solution. You will realise that exactly these were the answers given by Appu, Sarita and Ameena. They had learnt to set up equations and solve them. That is why they could construct their mind reader game and impress the whole class.
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Now let us tackle Exercise 4.2. Question 1: Give first the step you will use to separate the variable and then solve the equation. (a) x – 1 = 0. Step: Add 1 to both sides. Solution: x = 1. (b) x + 1 = 0. Step: Subtract 1 from both sides. Solution: x = –1. (c) x – 1 = 5. Step: Add 1 to both sides. Solution: x = 6. (d) x + 6 = 2. Step: Subtract 6 from both sides. Solution: x = –4. (e) y – 4 = –7. Step: Add 4 to both sides. Solution: y = –3. (f) y – 4 = 4. Step: Add 4 to both sides. Solution: y = 8. (g) y + 4 = 4. Step: Subtract 4 from both sides. Solution: y = 0. (h) y + 4 = –4. Step: Subtract 4 from both sides. Solution: y = –8.
Question 2: Give first the step you will use to separate the variable and then solve the equation. (a) 3l = 42. Step: Divide both sides by 3. Solution: l = 14. (b) b/2 = 6. Step: Multiply both sides by 2. Solution: b = 12. (c) p/7 = 4. Step: Multiply both sides by 7. Solution: p = 28. (d) 4x = 25. Step: Divide both sides by 4. Solution: x = 25/4. (e) 8y = 36. Step: Divide both sides by 8. Solution: y = 36/8, which simplifies to 9/2. (f) z/3 = 5/4. Step: Multiply both sides by 3. Solution: z = 15/4. (g) a/5 = 7/15. Step: Multiply both sides by 5. Solution: a = 35/15, which simplifies to 7/3. (h) 20t = –10. Step: Divide both sides by 20. Solution: t = –10/20, which simplifies to –1/2.
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Question 3: Give the steps you will use to separate the variable and then solve the equation. (a) 3n – 2 = 46. Step one: Add 2 to both sides to get 3n = 48. Step two: Divide both sides by 3 to get n = 16. (b) 5m + 7 = 17. Step one: Subtract 7 from both sides to get 5m = 10. Step two: Divide both sides by 5 to get m = 2. (c) 20p/3 = 40. Step one: Multiply both sides by 3 to get 20p = 120. Step two: Divide both sides by 20 to get p = 6. (d) 3p/10 = 6. Step one: Multiply both sides by 10 to get 3p = 60. Step two: Divide both sides by 3 to get p = 20.
Question 4: Solve the following equations. (a) 10p = 100. Divide by 10: p = 10. (b) 10p + 10 = 100. Subtract 10: 10p = 90. Divide by 10: p = 9. (c) p/4 = 5. Multiply by 4: p = 20. (d) –p/3 = 5. Multiply by 3: –p = 15. Multiply by –1: p = –15. (e) 3p/4 = 6. Multiply by 4: 3p = 24. Divide by 3: p = 8. (f) 3s = –9. Divide by 3: s = –3. (g) 3s + 12 = 0. Subtract 12: 3s = –12. Divide by 3: s = –4. (h) 3s = 0. Divide by 3: s = 0. (i) 2q = 6. Divide by 2: q = 3. (j) 2q – 6 = 0. Add 6: 2q = 6. Divide by 2: q = 3. (k) 2q + 6 = 0. Subtract 6: 2q = –6. Divide by 2: q = –3. (l) 2q + 6 = 12. Subtract 6: 2q = 6. Divide by 2: q = 3.
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Let us practise solving some more equations. While solving these equations, we shall learn about transposing a number, that is, moving it from one side to the other. We can transpose a number instead of adding or subtracting it from both sides of the equation. Example 6. Solve 12p – 5 = 25. We will call this equation 4.12. Solution: Adding 5 on both sides of the equation, 12p – 5 + 5 = 25 + 5, or 12p = 30. Dividing both sides by 12, 12p / 12 = 30 / 12, or p = 5/2. Check: Putting p = 5/2 in the LHS of the equation, LHS = 12 × 5/2 – 5 = 6 × 5 – 5 = 30 – 5 = 25, which equals RHS. Note, adding 5 to both sides is the same as changing side of –5. Changing side is called transposing. While transposing a number, we change its sign. As we have seen, while solving equations one commonly used operation is adding or subtracting the same number on both sides of the equation. Transposing a number is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed. What applies to numbers also applies to expressions.
Let us take two more examples of transposing. First, 3p – 10 = 5. Transpose –10 from LHS to RHS. On transposing –10 becomes +10. So 3p = 5 + 10 or 3p = 15. Second, 5x + 12 = 27. Transposing +12. On transposing +12 becomes –12. So 5x = 27 – 12 or 5x = 15.
We shall now solve two more equations. As you can see they involve brackets, which have to be solved before proceeding. Example 7. Solve part (a): 4(m + 3) = 18. Solution: Let us divide both the sides by 4. This will remove the brackets in the LHS. We get m + 3 = 18/4 or m + 3 = 9/2. Now m = 9/2 – 3, by transposing 3 to RHS. Or m = 3/2. This is the required solution. Check: LHS = 4(3/2 + 3). Put m = 3/2. This equals 4 × 3/2 + 4 × 3, which is 6 + 12 = 18, which equals RHS. Part (b): –2(x + 3) = 8. Solution: We divide both sides by –2, so as to remove the brackets in the LHS. We get x + 3 = 8 / –2 or x + 3 = –4. That is, x = –4 – 3, by transposing 3 to RHS. Or x = –7. This is the required solution. Check: LHS = –2(–7 + 3) = –2(–4) = 8, which equals RHS as required.
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Now let us move to applications of simple equations to practical situations. We have already seen examples in which we have taken statements in everyday language and converted them into simple equations. We also have learnt how to solve simple equations. Thus we are ready to solve puzzles and problems from practical situations. The method is first to form equations corresponding to such situations and then to solve those equations to give the solution to the puzzles and problems. Example 8. The sum of three times a number and 11 is 32. Find the number. Solution: If the unknown number is taken to be x, then three times the number is 3x and the sum of 3x and 11 is 32. That is, 3x + 11 = 32. To solve this equation, we transpose 11 to RHS, so that 3x = 32 – 11 or 3x = 21. Now, divide both sides by 3. So x = 21 / 3 = 7. The required number is 7. We may check it by taking 3 times 7 and adding 11 to it. It gives 32 as required.
Example 9. Find a number, such that one-fourth of the number is 3 more than 7. Solution: Let us take the unknown number to be y; one-fourth of y is y/4. This number is more than 7 by 3. Hence we get the equation for y as y/4 – 7 = 3. To solve this equation, first transpose 7 to RHS. We get y/4 = 3 + 7 = 10. We then multiply both sides of the equation by 4, to get y/4 × 4 = 10 × 4 or y = 40. The required number is 40. Let us check the equation formed. Putting the value of y in the equation, LHS = 40/4 – 7 = 10 – 7 = 3, which equals RHS, as required.
Try These: (i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is? Let the number be n. Equation: 6n – 5 = 7. Add 5: 6n = 12. Divide by 6: n = 2. The number is 2. (ii) What is that number one third of which added to 5 gives 8? Let the number be k. Equation: k/3 + 5 = 8. Subtract 5: k/3 = 3. Multiply by 3: k = 9. The number is 9.
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Example 10. Raju’s father’s age is 5 years more than three times Raju’s age. Find Raju’s age, if his father is 44 years old. Solution: As given earlier, the equation that gives Raju's age is 3y + 5 = 44. To solve it, we first transpose 5, to get 3y = 44 – 5 = 39. Dividing both sides by 3, we get y = 13. That is, Raju’s age is 13 years. You may check the answer.
Try These: There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box? Let the smaller box contain m mangoes. Equation: 8m + 4 = 100. Subtract 4: 8m = 96. Divide by 8: m = 12. Each small box contains 12 mangoes.
Now let us solve Exercise 4.3. Question 1: Set up equations and solve them to find the unknown numbers in the following cases. (a) Add 4 to eight times a number; you get 60. Let the number be x. Equation: 8x + 4 = 60. Subtract 4: 8x = 56. Divide by 8: x = 7. (b) One-fifth of a number minus 4 gives 3. Let the number be y. Equation: y/5 – 4 = 3. Add 4: y/5 = 7. Multiply by 5: y = 35. (c) If I take three-fourths of a number and add 3 to it, I get 21. Let the number be z. Equation: 3z/4 + 3 = 21. Subtract 3: 3z/4 = 18. Multiply by 4: 3z = 72. Divide by 3: z = 24. (d) When I subtracted 11 from twice a number, the result was 15. Let the number be a. Equation: 2a – 11 = 15. Add 11: 2a = 26. Divide by 2: a = 13. (e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8. Let the number of notebooks be n. Equation: 50 – 3n = 8. Subtract 50: –3n = –42. Divide by –3: n = 14. (f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8. Let the number be b. Equation: (b + 19)/5 = 8. Multiply by 5: b + 19 = 40. Subtract 19: b = 21. (g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23. Let the number be c. Equation: 5c/2 – 7 = 23. Add 7: 5c/2 = 30. Multiply by 2: 5c = 60. Divide by 5: c = 12.
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Question 2: Solve the following. (a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score? Let lowest score be l. Equation: 2l + 7 = 87. Subtract 7: 2l = 80. Divide by 2: l = 40. The lowest score is 40. (b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? Remember, the sum of three angles of a triangle is 180°. Let each base angle be b. Equation: b + b + 40 = 180. 2b + 40 = 180. Subtract 40: 2b = 140. Divide by 2: b = 70. The base angles are 70° each. (c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score? Let Rahul's runs be r. Sachin's runs are 2r. Together they fell two short of 200, so their sum is 198. Equation: r + 2r = 198. 3r = 198. Divide by 3: r = 66. Rahul scored 66 runs. Sachin scored 2 × 66, which is 132 runs.
Question 3: Solve the following. (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have? Let Parmit's marbles be m. Equation: 5m + 7 = 37. Subtract 7: 5m = 30. Divide by 5: m = 6. Parmit has 6 marbles. (ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi's age? Let Laxmi's age be y. Equation: 3y + 4 = 49. Subtract 4: 3y = 45. Divide by 3: y = 15. Laxmi is 15 years old. (iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77? Let fruit trees be f. Equation: 3f + 2 = 77. Subtract 2: 3f = 75. Divide by 3: f = 25. Twenty-five fruit trees were planted.
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Question 4: Solve the following riddle: I am a number, Tell my identity! Take me seven times over And add a fifty! To reach a triple century You still need forty! Let the number be x. Seven times the number is 7x. Add fifty: 7x + 50. To reach a triple century, which is 300, you still need 40. So 7x + 50 + 40 = 300. Or 7x + 50 = 260. Subtract 50: 7x = 210. Divide by 7: x = 30. The number is 30.
Let us review what we have discussed in this chapter under What Have We Discussed? One, an equation is a condition on a variable such that two expressions in the variable should have equal value. Two, the value of the variable for which the equation is satisfied is called the solution of the equation. Three, an equation remains the same if the LHS and the RHS are interchanged. Four, in case of the balanced equation, if we add the same number to both the sides, or subtract the same number from both the sides, or multiply both sides by the same number, or divide both sides by the same number, the balance remains undisturbed, that is, the value of the LHS remains equal to the value of the RHS. Five, the above property gives a systematic method of solving an equation. We carry out a series of identical mathematical operations on the two sides of the equation in such a way that on one of the sides we get just the variable. The last step is the solution of the equation. Six, transposing means moving to the other side. Transposition of a number has the same effect as adding same number to or subtracting the same number from both sides of the equation. When you transpose a number from one side of the equation to the other side, you change its sign. For example, transposing +3 from the LHS to the RHS in equation x + 3 = 8 gives x = 8 – 3, which is 5. We can carry out the transposition of an expression in the same way as the transposition of a number. Seven, we also learnt how, using the technique of doing the same mathematical operation, for example adding the same number, on both sides, we could build an equation starting from its solution. Further, we also learnt that we could relate a given equation to some appropriate practical situation and build a practical word problem or puzzle from the equation.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]