Welcome dear students! Today we are going to learn about Linear Equations in One Variable from Class 8 Maths.
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, 2y + 5/2 = 37/2, 6z + 10 = -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: 2x, 2x + 1, 3y - 7, 12 - 5z, 5/4(x - 4) + 10. These are not linear expressions: x^2 + 1, y + y^2, 1 + z + z^2 + z^3, since highest power of variable is greater than 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.
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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, or LHS. The expression on the right of the equality sign is the Right Hand Side, or RHS. In the diagram provided in your textbook, we see the equation 2x - 3 = 7 clearly labeled. The variable is x, the equality sign sits in the center, the entire statement is the equation, 2x - 3 is marked as the LHS, and 7 is marked as the RHS. 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. For instance, x = 5 is the solution of the equation 2x - 3 = 7. For x = 5, LHS = 2 × 5 - 3 = 7, which equals 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. 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.
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Now let us move on to section 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.
Let us look at Example 1. Solve 2x - 3 = x + 2. Solution: We have 2x = x + 2 + 3 or 2x = x + 5 or 2x - x = x + 5 - x, which means subtracting x from both sides, or x = 5, which is the solution. Here we subtracted from both sides of the equation, not a number or 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.
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Now, Example 2. Solve 5x + 7/2 = 3/2 x - 14. Solution: Multiply both sides of the equation by 2. We get 2 × (5x + 7/2) = 2 × (3/2 x - 14). This gives (2 × 5x) + (2 × 7/2) = (2 × 3/2 x) - (2 × 14), or 10x + 7 = 3x - 28, or 10x - 3x + 7 = -28, which is transposing 3x to LHS, or 7x + 7 = -28, or 7x = -28 - 7, or 7x = -35, or x = -35/7, or x = -5, which is the solution.
Let us now practice with Exercise 2.1. I will solve each equation and check the results for you. Question 1: 3x = 2x + 18. Subtract 2x from both sides: 3x - 2x = 18, so x = 18. Check: LHS = 3 × 18 = 54. RHS = 2 × 18 + 18 = 36 + 18 = 54. LHS equals RHS. Question 2: 5t - 3 = 3t - 5. Subtract 3t from both sides and add 3 to both sides: 5t - 3t = -5 + 3, so 2t = -2, t = -1. Check: LHS = 5 × -1 - 3 = -5 - 3 = -8. RHS = 3 × -1 - 5 = -3 - 5 = -8. LHS equals RHS. Question 3: 5x + 9 = 5 + 3x. Subtract 3x from both sides and subtract 9 from both sides: 5x - 3x = 5 - 9, so 2x = -4, x = -2. Check: LHS = 5 × -2 + 9 = -10 + 9 = -1. RHS = 5 + 3 × -2 = 5 - 6 = -1. LHS equals RHS. Question 4: 4z + 3 = 6 + 2z. Subtract 2z and 3 from both sides: 4z - 2z = 6 - 3, so 2z = 3, z = 3/2. Check: LHS = 4 × 3/2 + 3 = 6 + 3 = 9. RHS = 6 + 2 × 3/2 = 6 + 3 = 9. LHS equals RHS.
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Question 5: 2x - 1 = 14 - x. Add x and 1 to both sides: 2x + x = 14 + 1, so 3x = 15, x = 5. Check: LHS = 2 × 5 - 1 = 9. RHS = 14 - 5 = 9. LHS equals RHS. Question 6: 8x + 4 = 3(x - 1) + 7. Open brackets on RHS: 8x + 4 = 3x - 3 + 7, so 8x + 4 = 3x + 4. Subtract 3x and 4 from both sides: 5x = 0, x = 0. Check: LHS = 0 + 4 = 4. RHS = 3(0 - 1) + 7 = -3 + 7 = 4. LHS equals RHS. Question 7: x = 4/5 (x + 10). Multiply both sides by 5: 5x = 4(x + 10), so 5x = 4x + 40. Subtract 4x: x = 40. Check: LHS = 40. RHS = 4/5 (40 + 10) = 4/5 × 50 = 40. LHS equals RHS. Question 8: 2x/3 + 1 = 7x/15 + 3. Multiply by 15: 15 × (2x/3) + 15 × 1 = 15 × (7x/15) + 15 × 3, so 10x + 15 = 7x + 45. Subtract 7x and 15: 3x = 30, x = 10. Check: LHS = 20/3 + 1 = 23/3. RHS = 70/15 + 3 = 14/3 + 9/3 = 23/3. LHS equals RHS. Question 9: 2y + 5/3 = 26/3 - y. Add y to both sides and subtract 5/3 from both sides: 3y = 26/3 - 5/3, so 3y = 21/3 = 7. y = 7/3. Check: LHS = 2 × 7/3 + 5/3 = 14/3 + 5/3 = 19/3. RHS = 26/3 - 7/3 = 19/3. LHS equals RHS. Question 10: 3m = 5m - 8/5. Subtract 5m: -2m = -8/5. Multiply by -1: 2m = 8/5. Divide by 2: m = 4/5. Check: LHS = 3 × 4/5 = 12/5. RHS = 5 × 4/5 - 8/5 = 20/5 - 8/5 = 12/5. LHS equals RHS.
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Now let us move to section 2.3, Reducing Equations to Simpler Form. Example 3: Solve (6x+1)/3 + 1 = (x-3)/6. Solution: Multiplying both sides of the equation by 6, we get 6 × (6x+1)/3 + 6 × 1 = 6 × (x-3)/6. This simplifies to 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, which is the required solution. Why 6? Because it is the smallest multiple or LCM of the given denominators. Let us check: LHS = (6(-1)+1)/3 + 1 = -5/3 + 1 = -5/3 + 3/3 = -2/3. RHS = ((-1)-3)/6 = -4/6 = -2/3. LHS equals RHS, as required.
Example 4: Solve 5x - 2(2x - 7) = 2(3x - 1) + 7/2. Solution: Let us open the brackets. LHS = 5x - 4x + 14 = x + 14. RHS = 6x - 2 + 7/2 = 6x - 4/2 + 7/2 = 6x + 3/2. The equation is x + 14 = 6x + 3/2. Transposing terms: 14 = 6x - x + 3/2, or 14 = 5x + 3/2. Transposing 3/2: 14 - 3/2 = 5x, or 28/2 - 3/2 = 5x, or 25/2 = 5x. Dividing by 5: x = 25/2 × 1/5 = 5/2. Therefore, required solution is x = 5/2. Check: LHS = 5 × 5/2 - 2(5/2 × 2 - 7) = 25/2 - 2(5 - 7) = 25/2 - 2(-2) = 25/2 + 4 = 25/2 + 8/2 = 33/2. RHS = 2(5/2 × 3 - 1) + 7/2 = 2(15/2 - 2/2) + 7/2 = 2 × 13/2 + 7/2 = 13 + 7/2 = 26/2 + 7/2 = 33/2. LHS equals RHS, as required. Did you observe how we simplified the form of the given equation? Here, we had to multiply both sides of the equation by the LCM of the denominators of the terms in the expressions of the equation. Note, in this example we brought the equation to a simpler form by opening brackets and combining like terms on both sides of the equation.
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Now let us solve Exercise 2.2. Question 1: x/2 - 1/5 = x/3 + 1/4. Multiply by 60: 30x - 12 = 20x + 15. 10x = 27, x = 27/10. Question 2: n/2 - 3n/4 + 5n/6 = 21. Multiply by 12: 6n - 9n + 10n = 252. 7n = 252, n = 36. Question 3: x + 7 - 8x/3 = 17/6 - 5x/2. Multiply by 6: 6x + 42 - 16x = 17 - 15x. -10x + 42 = 17 - 15x. Add 15x and subtract 42: 5x = -25, x = -5. Question 4: (x-5)/3 = (x-3)/5. Multiply by 15: 5(x-5) = 3(x-3). 5x - 25 = 3x - 9. 2x = 16, x = 8. Question 5: (3t-2)/4 - (2t+3)/3 = 2/3 - t. Multiply by 12: 3(3t-2) - 4(2t+3) = 8 - 12t. 9t - 6 - 8t - 12 = 8 - 12t. t - 18 = 8 - 12t. 13t = 26, t = 2. Question 6: m - (m-1)/2 = 1 - (m-2)/3. Multiply by 6: 6m - 3(m-1) = 6 - 2(m-2). 6m - 3m + 3 = 6 - 2m + 4. 3m + 3 = 10 - 2m. 5m = 7, m = 7/5.
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Question 7: 3(t - 3) = 5(2t + 1). 3t - 9 = 10t + 5. -7t = 14, t = -2. Question 8: 15(y - 4) - 2(y - 9) + 5(y + 6) = 0. 15y - 60 - 2y + 18 + 5y + 30 = 0. 18y - 12 = 0. 18y = 12, y = 12/18 = 2/3. Question 9: 3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17. 15z - 21 - 18z + 22 = 32z - 52 - 17. -3z + 1 = 32z - 69. -35z = -70, z = 2. Question 10: 0.25(4f - 3) = 0.05(10f - 9). Multiply by 100: 25(4f - 3) = 5(10f - 9). Divide by 5: 5(4f - 3) = 10f - 9. 20f - 15 = 10f - 9. 10f = 6, f = 6/10 = 3/5.
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Finally, let us review what we have discussed. First, 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. Second, the equations we study in Classes 6, 7 and 8 are linear equations in one variable. In such equations, the expressions which form the equation contain only one variable. Further, the equations are linear, meaning the highest power of the variable appearing in the equation is 1. Third, an equation may have linear expressions on both sides. Equations that we studied in Classes 6 and 7 had just a number on one side of the equation. Fourth, just as numbers, variables can also be transposed from one side of the equation to the other. Fifth, 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. Sixth, the utility of linear equations is in their diverse applications; different problems on numbers, ages, perimeters, combination of currency notes, and various other practical situations can be solved using linear equations.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]