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       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, x2 + 1, y + y2 5 37 Some examples of equations are: 5x = 25, 2x – 3 = 9, 2 y + = , 6 z + 10 = −2 2 2 You 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: 5 2x, 2x + 1, 3y – 7, 12 – 5z, ( x – 4) + 10 4 These are not linear expressions: x2 + 1, y + y2, 1 + z + z2 + z3 (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 2x – 3 = 7 on the right of the equality sign is the Right 2x – 3 = LHS Hand Side (RHS). 7 = RHS Reprint 2024-25   (b) In an equation the values of x = 5 is the solution of the equation the expressions on the LHS 2x – 3 = 7. For x = 5, and RHS are equal. This LHS = 2 × 5 – 3 = 7 = RHS happens to be true only for On the other hand x = 10 is not a solution of the certain values of the variable. equation. For x = 10, LHS = 2 × 10 – 3 = 17. These values are the This is not equal to the RHS solutions of the equation. (c) How to find the solution of an equation? We 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 having the Variable on both Sides An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7, the two expressions are 2x – 3 and 7. In most examples that we have come across so far, the RHS is just a number. But this need not always be so; both sides could have expressions with variables. For example, the equation 2x – 3 = x + 2 has expressions with a variable on both sides; the expression on the LHS is (2x – 3) and the expression on the RHS is (x + 2). We now discuss how to solve such equations which have expressions with the variable on both sides. Example 1: Solve 2x – 3 = x + 2 Solution: We have 2x = x + 2 + 3 or 2x = x + 5 or 2x – x = x + 5 – x (subtracting x from both sides) or x=5 (solution) Here we subtracted from both sides of the equation, not a number (constant), but a term involving the variable. We can do this as variables are also numbers. Also, note that subtracting x from both sides amounts to transposing x to LHS. 7 3 Example 2: Solve 5x + = x − 14 2 2 Solution: Multiply both sides of the equation by 2. We get  7 3  2 ×  5x +  = 2 ×  x − 14  2 2 Reprint 2024-25     7  3  (2 × 5x) +  2 ×  =  2 × x − (2 × 14) 2 2 or 10x + 7 = 3x – 28 or 10x – 3x + 7 = – 28 (transposing 3x to LHS) or 7x + 7 = – 28 or 7x = – 28 – 7 or 7x = – 35 −35 or x= or x=–5 (solution) 7 EXERCISE 2.1 Solve the following equations and check your results. 1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x 4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3 (x – 1) + 7 4 2x 7x 5 26 7. x = (x + 10) 8. +1= +3 9. 2y + = −y 5 3 15 3 3 8 10. 3m = 5 m – 5 2.3 Reducing Equations to Simpler Form 6x + 1 x−3 Example 16: Solve +1 = 3 6 Why 6? Because it is the smallest multiple (or LCM) Solution: Multiplying both sides of the equation by 6, of the given denominators. 6 (6 x + 1) 6( x − 3) + 6 ×1 = 3 6 or 2 (6x + 1) + 6 = x – 3 or 12x + 2 + 6 = x – 3 (opening the brackets ) or 12x + 8 = x – 3 or 12x – x + 8 = – 3 or 11x + 8 = – 3 or 11x = –3 – 8 or 11x = –11 or x =–1 (required solution) Reprint 2024-25   6(−1) + 1 −6 + 1 −5 3 −5 + 3 −2 Check: LHS = +1 = +1 = + = = 3 3 3 3 3 3 ( −1) − 3 −4 −2 RHS = = = 6 6 3 LHS = RHS. (as required) 7 Example 17: Solve 5x – 2 (2x – 7) = 2 (3x – 1) + 2 Solution: Let us open the brackets, LHS = 5x – 4x + 14 = x + 14 7 4 7 3 RHS = 6x – 2 + = 6x − + = 6x + 2 2 2 2 3 The equation is x + 14 = 6x + 2 3 or 14 = 6x – x + 2 3 or 14 = 5x + 2 3 3 or 14 – = 5x (transposing ) 2 2 28 − 3 or = 5x 2 Did you observe how we simplified the form of the given 25 equation? Here, we had to or = 5x 2 multiply both sides of the equation by the LCM of the 25 1 5 × 5 5 or x= × = = denominators of the terms in the 2 5 2×5 2 expressions of the equation. 5 Therefore, required solution is x =. 2 Check: LHS = 25 25 25 25 + 8 33 Note, in this example we = − 2(5 − 7) = − 2( −2) = +4 = = brought the equation to a 2 2 2 2 2 simpler form by opening brackets and combining like terms on both sides of the RHS = equation. 26 + 7 33 = = = LHS. (as required) 2 2 Reprint 2024-25    EXERCISE 2.2 Solve the following linear equations. x 1 x 1 n 3n 5n 8 x 17 5 x 1. − = + 2. − + = 21 3. x + 7 − = − 2 5 3 4 2 4 6 3 6 2 x−5 x−3 3t − 2 2t + 3 2 m −1 m−2 4. = 5. − = −t 6. m − =1− 3 5 4 3 3 2 3 Simplify and solve the following linear equations. 7. 3(t – 3) = 5(2t + 1) 8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0 9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 10. 0.25(4f – 3) = 0.05(10f – 9) WHAT HAVE WE DISCUSSED? 1. An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side. 2. The equations we study in Classes VI, VII and VIII are linear equations in one variable. In such equations, the expressions which form the equation contain only one variable. Further, the equations are linear, i.e., the highest power of the variable appearing in the equation is 1. 3. An equation may have linear expressions on both sides. Equations that we studied in Classes VI and VII had just a number on one side of the equation. 4. Just as numbers, variables can, also, be transposed from one side of the equation to the other. 5. Occasionally, the expressions forming equations have to be simplified before we can solve them by usual methods. Some equations may not even be linear to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression. 6. The utility of linear equations is in their diverse applications; different problems on numbers, ages, perimeters, combination of currency notes, and so on can be solved using linear equations. Reprint 2024-25   NOTES Reprint 2024-25