SIMPLE EQUATIONS 4.4 WHAT EQUATION IS? 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. 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. 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 (i) Three times x is 3x. Sum of 3x and 11 is 3x + 11. The sum is 32. The equation is 3x + 11 = 32. (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 SIMPLE EQUATIONS EXERCISE 4.1 1. Complete the last column of the table. 2. Check whether the value given in the brackets is a solution to the given equation or not: (a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = – 2) (c) 7n + 5 = 19 (n = 2) (d) 4p – 3 = 13 (p = 1) (e) 4p – 3 = 13 (p = – 4) (f) 4p – 3 = 13 (p = 0) 3. Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4 4. Write equations for the following statements: (i) The sum of numbers x and 4 is 9. (ii) 2 subtracted from y is 8. (iii) Ten times a is 70. (iv) The number b divided by 5 gives 6. (v) Three-fourth of t is 15. (vi) Seven times m plus 7 gets you 77. (vii) One-fourth of a number x minus 4 gives 4. (viii) If you take away 6 from 6 times y, you get 60. (ix) If you add 3 to one-third of z, you get 30. 5. Write the following equations in statement forms: S. No. Equation Value Say, whether the Equation is Satisfied. (Yes/ No) (i) x + 3 = 0 x = 3 (ii) x + 3 = 0 x = 0 (iii) x + 3 = 0 x = – 3 (iv) x – 7 = 1 x = 7 (v) x – 7 = 1 x = 8 (vi) 5x = 25 x = 0 (vii) 5x = 25 x = 5 (viii) 5x = 25 x = – 5 (ix) m 3 = 2 m = – 6 (x) m 3 = 2 m = 0 (xi) m 3 = 2 m = 6 m(i) p + 4 = 15 (ii) m – 7 = 3 (iii) 2m = 7 (iv) = 353mp(v) = 6 (vi) 3p + 4 = 25 (vii) 4p – 2 = 18 (viii) + 2 = 85 2MATHEMATICS 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.) (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.) (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.) (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). 4.4.1 Solving an Equation Consider an equality 8 – 3 = 4 + 1 (4.5) The equality (4.5) 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 (i.e., its LHS and RHS are equal). 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. • 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. 5LHS = (8 – 3) ÷ 2 = 5 ÷ 2 = 2 5RHS = (4+1) ÷ 2 = 5 ÷ 2 = = LHS2Again, the equality holds. If we take any other equality, we shall find the same conclusions. Suppose, we do not observe these rules. Specificially, suppose we add different numbers, to the two sides of an equality. We shall find in this case that the equality does not MATHEMATICS 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 (i.e., subtracting 3) on both the sides of the equation, we arrived at the solution of the equation. • Let us look at another equation x – 3 = 10 (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 (4.7) 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 (4.8) m = 5 (4.9)2In the first case, we shall divide both the sides by 5. This will give us just y on LHS 5y 5× y 35 ×57New LHS = == y , New RHS = == 7 55 55 Therefore, y =7 This is the required solution. We can substitute y = 7 in Eq. (4.8) and check that it issatisfied. 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 dependson 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.Let us solve some more equations with this in mind. EXAMPLE 5 Solve: (a) 3n + 7 = 25 (4.10) (b) 2p – 1 = 23 (4.11) SOLUTION (a) 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 stepwe shall divide by 3 to get n. Remember we must do the same operation on bothsides of the equation. Therefore, subtracting 7 from both sides, 3n + 7 – 7 =25 – 7 (Step 1) or 3n =18 SIMPLE EQUATIONS Now divide both sides by 3, 3n18 = (Step 2)33 or n = 6, which is the solution. (b) What should we do here? First we shall add 1 to both the sides: 2p – 1 + 1 =23 + 1 (Step 1) or 2p =24 2 p 24Now divide both sides by 2, we get = (Step 2)22 or p = 12, which is the solution. One good practice you should develop is to check the solution you have obtained. Although we have not done this for (a) above, let us do it for this example. Let us put the solution p = 12 back into the equation. LHS = 2p – 1 = 2 × 12 – 1 = 24 – 1 = 23 = RHS The solution is thus checked for its correctness. Why do you not check the solution of (a) also? 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 the equations (4.1) and (4.2) which correspond respectively to Ameena’s and Appu’s examples. • First consider the equation 4x + 5 = 65. (4.1) Subtracting 5 from both sides, 4x + 5 – 5 = 65 – 5. i.e. 4x =60 4x60Divide both sides by 4; this will separate x. We get = 44 or x = 15, which is the solution. (Check, if it is correct.) • Now consider,10y – 20 = 50 (4.2) Adding 20 to both sides, we get 10y – 20 + 20 = 50 + 20 or 10y = 70 10y70Dividing both sides by 10, we get = 10 10 or y = 7, which is the solution. (Check if it is correct.) You will realise that exactly these were the answers given byAppu, Sarita andAmeena. 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. We shall come back to this in Section 4.7. SIMPLE EQUATIONS 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 (i.e., changing the side of the 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. Adding or Subtracting on both sides Transposing (i) 3p – 10 = 5 Add 10 to both sides (i) 3p – 10 = 5 Transpose (–10) from LHS to RHS or (ii) 3p – 10 + 10 = 5 + 10 3p = 15 5x + 12 = 27 Subtract 12 from both sides 5x + 12 – 12 = 27 – 12 (ii) (On transposing – 10 becomes + 10). 3p = 5 + 10 or 3p = 15 5x + 12 = 27 Transposing + 12 (On transposing + 12 becomes – 12) 5x = 27 – 12 or 5x = 15 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 (a) 4(m + 3) = 18 (b) – 2(x + 3) = 8 SOLUTION (a) 4(m + 3) = 18 Let us divide both the sides by 4. This will remove the brackets in the LHS We get, 18 9 m +=3 or m +=3 42 9 or m = –3 (transposing 3 to RHS)2 3 ⎛ 9 963⎞ as −= −=or m = (required solution) ⎜ 3 ⎟2 ⎝2 222⎠⎡3 ⎤ 33Check LHS = 4 + 4 2343 ]3 =× +×=×+×43 [put m = ⎢⎥ 2⎣2 ⎦2 = 6 + 12 = 18 = RHS (b) –2(x + 3) = 8 We divide both sides by (– 2), so as to remove the brackets in the LHS, we get, 8 x + 3 = – or x + 3 = – 42 i.e., x = – 4 – 3 (transposing 3 to RHS) or x = –7 (required solution) MATHEMATICS WHAT HAVE WE DISCUSSED? 1. An equation is a condition on a variable such that two expressions in the variable should have equal value. 2. The value of the variable for which the equation is satisfied is called the solution of the equation. 3. An equation remains the same if the LHS and the RHS are interchanged. 4. In case of the balanced equation, if we (i) add the same number to both the sides, or (ii) subtract the same number from both the sides, or (iii) multiply both sides by the same number, or (iv) divide both sides by the same number, the balance remains undisturbed, i.e., the value of the LHS remains equal to the value of the RHS 5. 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. 6. 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 (= 5). We can carry out the transposition of an expression in the same way as the transposition of a number. 7. We have learnt how to construct simple algebraic expressions corresponding to practical situations. 8. 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/puzzle from the equation.