Page 1 LINEAR EQUATIONS IN ONE VARIABLE 21 2.1 Introduction In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are: 5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x 2 + 1, y + y 2 Some examples of equations are: 5x = 25, 2x – 3 = 9, 5 37 2 , 6 10 2 2 2 y z + = + =  Y ou would remember that equations use the equality (=) sign; it is missing in expressions. Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover , the expressions we use to form equations are linear . This means that the highest power of the variable appearing in the expression is 1. These are linear expressions: 2x, 2x + 1, 3y – 7, 12 – 5z, 5 ( – 4) 10 4 x + These are not linear expressions: x 2 + 1, y + y 2 , 1 + z + z 2 + z 3 (since highest power of variable > 1) Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type. Let us briefly revise what we know: (a) An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). Linear Equations in One Variable CHAPTER 2 2x – 3 = 7 2x – 3 = LHS 7 = RHS 201920 Page 2 LINEAR EQUATIONS IN ONE VARIABLE 21 2.1 Introduction In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are: 5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x 2 + 1, y + y 2 Some examples of equations are: 5x = 25, 2x – 3 = 9, 5 37 2 , 6 10 2 2 2 y z + = + =  Y ou would remember that equations use the equality (=) sign; it is missing in expressions. Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover , the expressions we use to form equations are linear . This means that the highest power of the variable appearing in the expression is 1. These are linear expressions: 2x, 2x + 1, 3y – 7, 12 – 5z, 5 ( – 4) 10 4 x + These are not linear expressions: x 2 + 1, y + y 2 , 1 + z + z 2 + z 3 (since highest power of variable > 1) Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type. Let us briefly revise what we know: (a) An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). Linear Equations in One Variable CHAPTER 2 2x – 3 = 7 2x – 3 = LHS 7 = RHS 201920 22 MATHEMATICS (b) In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation. (c) How to find the solution of an equation? W e assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. A few such steps give the solution. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other Side Let us recall the technique of solving equations with some examples. Observe the solutions; they can be any rational number. Example 1: Find the solution of 2x – 3 = 7 Solution: Step 1 Add 3 to both sides. 2x – 3 + 3 = 7 + 3 (The balance is not disturbed) or 2x = 10 Step 2 Next divide both sides by 2. 2 2 x = 10 2 or x = 5 (required solution) Example 2: Solve 2y + 9 = 4 Solution: Transposing 9 to RHS 2y = 4 – 9 or 2y = – 5 Dividing both sides by 2, y = 5 2  (solution) To check the answer: LHS = 2  ? ? ? ? ? ? 5 2 + 9 = – 5 + 9 = 4 = RHS (as required) Do you notice that the solution  ? ? ? ? ? ? 5 2 is a rational number? In Class VII, the equations we solved did not have such solutions. x = 5 is the solution of the equation 2x – 3 = 7. For x = 5, LHS = 2 × 5 – 3 = 7 = RHS On the other hand x = 10 is not a solution of the equation. For x = 10, LHS = 2 × 10 –3 = 17. This is not equal to the RHS 201920 Page 3 LINEAR EQUATIONS IN ONE VARIABLE 21 2.1 Introduction In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are: 5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x 2 + 1, y + y 2 Some examples of equations are: 5x = 25, 2x – 3 = 9, 5 37 2 , 6 10 2 2 2 y z + = + =  Y ou would remember that equations use the equality (=) sign; it is missing in expressions. Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover , the expressions we use to form equations are linear . This means that the highest power of the variable appearing in the expression is 1. These are linear expressions: 2x, 2x + 1, 3y – 7, 12 – 5z, 5 ( – 4) 10 4 x + These are not linear expressions: x 2 + 1, y + y 2 , 1 + z + z 2 + z 3 (since highest power of variable > 1) Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type. Let us briefly revise what we know: (a) An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). Linear Equations in One Variable CHAPTER 2 2x – 3 = 7 2x – 3 = LHS 7 = RHS 201920 22 MATHEMATICS (b) In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation. (c) How to find the solution of an equation? W e assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. A few such steps give the solution. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other Side Let us recall the technique of solving equations with some examples. Observe the solutions; they can be any rational number. Example 1: Find the solution of 2x – 3 = 7 Solution: Step 1 Add 3 to both sides. 2x – 3 + 3 = 7 + 3 (The balance is not disturbed) or 2x = 10 Step 2 Next divide both sides by 2. 2 2 x = 10 2 or x = 5 (required solution) Example 2: Solve 2y + 9 = 4 Solution: Transposing 9 to RHS 2y = 4 – 9 or 2y = – 5 Dividing both sides by 2, y = 5 2  (solution) To check the answer: LHS = 2  ? ? ? ? ? ? 5 2 + 9 = – 5 + 9 = 4 = RHS (as required) Do you notice that the solution  ? ? ? ? ? ? 5 2 is a rational number? In Class VII, the equations we solved did not have such solutions. x = 5 is the solution of the equation 2x – 3 = 7. For x = 5, LHS = 2 × 5 – 3 = 7 = RHS On the other hand x = 10 is not a solution of the equation. For x = 10, LHS = 2 × 10 –3 = 17. This is not equal to the RHS 201920 LINEAR EQUATIONS IN ONE VARIABLE 23 Example 3: Solve 5 3 2 x + = 3 2  Solution: Transposing 5 2 to the RHS, we get 3 x = 3 5 8 2 2 2   = or 3 x = – 4 Multiply both sides by 3, x = – 4 × 3 or x = – 12 (solution) Check: LHS = 12 5 5 8 5 3 4 3 2 2 2 2  +   + =  + = = = RHS (as required) Do you now see that the coefficient of a variable in an equation need not be an integer? Example 4: Solve 15 4 – 7x = 9 Solution: We have 15 4 – 7x = 9 or – 7x = 9 – 15 4 (transposing 15 4 to R H S) or – 7x = 21 4 or x = 21 4 ( 7) ×  (dividing both sides by – 7) or x = 3 7 4 7 ×  × or x = 3 4  (solution) Check: LHS = 15 4 7 3 4   ? ? ? ? ? ? = 15 21 36 9 4 4 4 + = = = RHS (as required) EXERCISE 2.1 Solve the following equations. 1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2 4. 3 17 7 7 x + = 5. 6x = 12 6. 10 5 t = 7. 2 18 3 x = 8. 1.6 = 1.5 y 9. 7x – 9 = 16 201920 Page 4 LINEAR EQUATIONS IN ONE VARIABLE 21 2.1 Introduction In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are: 5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x 2 + 1, y + y 2 Some examples of equations are: 5x = 25, 2x – 3 = 9, 5 37 2 , 6 10 2 2 2 y z + = + =  Y ou would remember that equations use the equality (=) sign; it is missing in expressions. Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover , the expressions we use to form equations are linear . This means that the highest power of the variable appearing in the expression is 1. These are linear expressions: 2x, 2x + 1, 3y – 7, 12 – 5z, 5 ( – 4) 10 4 x + These are not linear expressions: x 2 + 1, y + y 2 , 1 + z + z 2 + z 3 (since highest power of variable > 1) Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type. Let us briefly revise what we know: (a) An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). Linear Equations in One Variable CHAPTER 2 2x – 3 = 7 2x – 3 = LHS 7 = RHS 201920 22 MATHEMATICS (b) In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation. (c) How to find the solution of an equation? W e assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. A few such steps give the solution. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other Side Let us recall the technique of solving equations with some examples. Observe the solutions; they can be any rational number. Example 1: Find the solution of 2x – 3 = 7 Solution: Step 1 Add 3 to both sides. 2x – 3 + 3 = 7 + 3 (The balance is not disturbed) or 2x = 10 Step 2 Next divide both sides by 2. 2 2 x = 10 2 or x = 5 (required solution) Example 2: Solve 2y + 9 = 4 Solution: Transposing 9 to RHS 2y = 4 – 9 or 2y = – 5 Dividing both sides by 2, y = 5 2  (solution) To check the answer: LHS = 2  ? ? ? ? ? ? 5 2 + 9 = – 5 + 9 = 4 = RHS (as required) Do you notice that the solution  ? ? ? ? ? ? 5 2 is a rational number? In Class VII, the equations we solved did not have such solutions. x = 5 is the solution of the equation 2x – 3 = 7. For x = 5, LHS = 2 × 5 – 3 = 7 = RHS On the other hand x = 10 is not a solution of the equation. For x = 10, LHS = 2 × 10 –3 = 17. This is not equal to the RHS 201920 LINEAR EQUATIONS IN ONE VARIABLE 23 Example 3: Solve 5 3 2 x + = 3 2  Solution: Transposing 5 2 to the RHS, we get 3 x = 3 5 8 2 2 2   = or 3 x = – 4 Multiply both sides by 3, x = – 4 × 3 or x = – 12 (solution) Check: LHS = 12 5 5 8 5 3 4 3 2 2 2 2  +   + =  + = = = RHS (as required) Do you now see that the coefficient of a variable in an equation need not be an integer? Example 4: Solve 15 4 – 7x = 9 Solution: We have 15 4 – 7x = 9 or – 7x = 9 – 15 4 (transposing 15 4 to R H S) or – 7x = 21 4 or x = 21 4 ( 7) ×  (dividing both sides by – 7) or x = 3 7 4 7 ×  × or x = 3 4  (solution) Check: LHS = 15 4 7 3 4   ? ? ? ? ? ? = 15 21 36 9 4 4 4 + = = = RHS (as required) EXERCISE 2.1 Solve the following equations. 1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2 4. 3 17 7 7 x + = 5. 6x = 12 6. 10 5 t = 7. 2 18 3 x = 8. 1.6 = 1.5 y 9. 7x – 9 = 16 201920 24 MATHEMATICS 10. 14y – 8 = 13 11. 17 + 6p = 9 12. 7 1 3 15 x + = 2.3 Some Applications W e begin with a simple example. Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers? We have a puzzle here. We do not know either of the two numbers, and we have to find them. W e are given two conditions. (i) One of the numbers is 10 more than the other. (ii) Their sum is 74. We already know from Class VII how to proceed. If the smaller number is taken to be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that the sum of these two numbers x and x + 10 is 74. This means that x + (x + 10) = 74. or 2x + 10 = 74 Transposing 10 to RHS, 2x = 74 – 10 or 2x = 64 Dividing both sides by 2, x = 32. This is one number. The other number is x + 10 = 32 + 10 = 42 The desired numbers are 32 and 42. (Their sum is indeed 74 as given and also one number is 10 more than the other.) W e shall now consider several examples to show how useful this method is. Example 5: What should be added to twice the rational number 7 3  to get 3 7 ? Solution: T wice the rational number 7 3  is 2 7 3 14 3 ×  ? ? ? ? ? ? =  . Suppose x added to this number gives 3 7 ; i.e., x +  ? ? ? ? ? ? 14 3 = 3 7 or 14 3 x  = 3 7 or x = 3 14 7 3 + (transposing 14 3 to RHS) = (3 3) (14 7) 21 × + × = 9 98 107 21 21 + = . 201920 Page 5 LINEAR EQUATIONS IN ONE VARIABLE 21 2.1 Introduction In the earlier classes, you have come across several algebraic expressions and equations. Some examples of expressions we have so far worked with are: 5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x 2 + 1, y + y 2 Some examples of equations are: 5x = 25, 2x – 3 = 9, 5 37 2 , 6 10 2 2 2 y z + = + =  Y ou would remember that equations use the equality (=) sign; it is missing in expressions. Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover , the expressions we use to form equations are linear . This means that the highest power of the variable appearing in the expression is 1. These are linear expressions: 2x, 2x + 1, 3y – 7, 12 – 5z, 5 ( – 4) 10 4 x + These are not linear expressions: x 2 + 1, y + y 2 , 1 + z + z 2 + z 3 (since highest power of variable > 1) Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type. Let us briefly revise what we know: (a) An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). Linear Equations in One Variable CHAPTER 2 2x – 3 = 7 2x – 3 = LHS 7 = RHS 201920 22 MATHEMATICS (b) In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation. (c) How to find the solution of an equation? W e assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. A few such steps give the solution. 2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other Side Let us recall the technique of solving equations with some examples. Observe the solutions; they can be any rational number. Example 1: Find the solution of 2x – 3 = 7 Solution: Step 1 Add 3 to both sides. 2x – 3 + 3 = 7 + 3 (The balance is not disturbed) or 2x = 10 Step 2 Next divide both sides by 2. 2 2 x = 10 2 or x = 5 (required solution) Example 2: Solve 2y + 9 = 4 Solution: Transposing 9 to RHS 2y = 4 – 9 or 2y = – 5 Dividing both sides by 2, y = 5 2  (solution) To check the answer: LHS = 2  ? ? ? ? ? ? 5 2 + 9 = – 5 + 9 = 4 = RHS (as required) Do you notice that the solution  ? ? ? ? ? ? 5 2 is a rational number? In Class VII, the equations we solved did not have such solutions. x = 5 is the solution of the equation 2x – 3 = 7. For x = 5, LHS = 2 × 5 – 3 = 7 = RHS On the other hand x = 10 is not a solution of the equation. For x = 10, LHS = 2 × 10 –3 = 17. This is not equal to the RHS 201920 LINEAR EQUATIONS IN ONE VARIABLE 23 Example 3: Solve 5 3 2 x + = 3 2  Solution: Transposing 5 2 to the RHS, we get 3 x = 3 5 8 2 2 2   = or 3 x = – 4 Multiply both sides by 3, x = – 4 × 3 or x = – 12 (solution) Check: LHS = 12 5 5 8 5 3 4 3 2 2 2 2  +   + =  + = = = RHS (as required) Do you now see that the coefficient of a variable in an equation need not be an integer? Example 4: Solve 15 4 – 7x = 9 Solution: We have 15 4 – 7x = 9 or – 7x = 9 – 15 4 (transposing 15 4 to R H S) or – 7x = 21 4 or x = 21 4 ( 7) ×  (dividing both sides by – 7) or x = 3 7 4 7 ×  × or x = 3 4  (solution) Check: LHS = 15 4 7 3 4   ? ? ? ? ? ? = 15 21 36 9 4 4 4 + = = = RHS (as required) EXERCISE 2.1 Solve the following equations. 1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2 4. 3 17 7 7 x + = 5. 6x = 12 6. 10 5 t = 7. 2 18 3 x = 8. 1.6 = 1.5 y 9. 7x – 9 = 16 201920 24 MATHEMATICS 10. 14y – 8 = 13 11. 17 + 6p = 9 12. 7 1 3 15 x + = 2.3 Some Applications W e begin with a simple example. Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers? We have a puzzle here. We do not know either of the two numbers, and we have to find them. W e are given two conditions. (i) One of the numbers is 10 more than the other. (ii) Their sum is 74. We already know from Class VII how to proceed. If the smaller number is taken to be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that the sum of these two numbers x and x + 10 is 74. This means that x + (x + 10) = 74. or 2x + 10 = 74 Transposing 10 to RHS, 2x = 74 – 10 or 2x = 64 Dividing both sides by 2, x = 32. This is one number. The other number is x + 10 = 32 + 10 = 42 The desired numbers are 32 and 42. (Their sum is indeed 74 as given and also one number is 10 more than the other.) W e shall now consider several examples to show how useful this method is. Example 5: What should be added to twice the rational number 7 3  to get 3 7 ? Solution: T wice the rational number 7 3  is 2 7 3 14 3 ×  ? ? ? ? ? ? =  . Suppose x added to this number gives 3 7 ; i.e., x +  ? ? ? ? ? ? 14 3 = 3 7 or 14 3 x  = 3 7 or x = 3 14 7 3 + (transposing 14 3 to RHS) = (3 3) (14 7) 21 × + × = 9 98 107 21 21 + = . 201920 LINEAR EQUATIONS IN ONE VARIABLE 25 Thus 107 21 should be added to 2 7 3 ×  ? ? ? ? ? ? to give 3 7 . Example 6: The perimeter of a rectangle is 13 cm and its width is 3 2 4 cm. Find its length. Solution: Assume the length of the rectangle to be x cm. The perimeter of the rectangle = 2 × (length + width) = 2 × (x + 3 2 4 ) = 2 11 4 x + ? ? ? ? ? ? The perimeter is given to be 13 cm. Therefore, 2 11 4 x + ? ? ? ? ? ? = 13 or 11 4 x + = 13 2 (dividing both sides by 2) or x = 13 11 2 4  = 26 11 15 3 3 4 4 4 4  = = The length of the rectangle is 3 3 4 cm. Example 7: The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages. Solution: Let Sahil’s present age be x years. It is given that this sum is 66 years. Therefore, 4x + 10 = 66 This equation determines Sahil’s present age which is x years. To solve the equation, We could also choose Sahil’s age 5 years later to be x and proceed. Why don’t you try it that way? Sahil Mother Sum Present age x 3x Age 5 years later x + 5 3x + 5 4x + 10 201920Read More
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