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Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > NCERT Solutions: Linear Equations in One Variable(Exercise 2.1 & 2.2)
NCERT Solutions for Class 8 Maths - Linear Equations in One Variable (Exercise 2.1)
Exercise 2.1
Q1: 3x = 2x + 18
Ans: 3x = 2x + 18
⇒ 3x – 2x = 18
⇒ x = 18
Putting the value of x in RHS and LHS, we get, 3 × 18 = (2 × 18) +18
⇒ 54 = 54
⇒ LHS = RHS
Q2: 5t – 3 = 3t – 5
Ans: 5t – 3 = 3t – 5
⇒ 5t – 3t = -5 + 3
⇒ 2t = -2
⇒ t = -1
Putting the value of t in RHS and LHS, we get, 5× (-1) – 3 = 3× (-1) – 5
⇒ -5 – 3 = -3 – 5
⇒ -8 = -8
⇒ LHS = RHS
Q3: 5x + 9 = 5 + 3x
Ans: 5x + 9 = 5 + 3x
⇒ 5x – 3x = 5 – 9
⇒ 2x = -4
⇒ x = -2
Putting the value of x in RHS and LHS, we get, 5× (-2) + 9 = 5 + 3× (-2)
⇒ -10 + 9 = 5 + (-6)
⇒ -1 = -1
⇒ LHS = RHS
Q4: 4z + 3 = 6 + 2z
Ans: 4z + 3 = 6 + 2z
⇒ 4z – 2z = 6 – 3
⇒ 2z = 3
⇒ z = 3/2
Putting the value of z in RHS and LHS, we get,
(4 × 3/2) + 3 = 6 + (2 × 3/2)
⇒ 6 + 3 = 6 + 3
⇒ 9 = 9
⇒ LHS = RHS
Q5: 2x – 1 = 14 – x
Ans: 2x – 1 = 14 – x
⇒ 2x + x = 14 + 1
⇒ 3x = 15
⇒ x = 5
Putting the value of x in RHS and LHS, we get, (2×5) – 1 = 14 – 5
⇒ 10 – 1 = 9
⇒ 9 = 9
⇒ LHS = RHS
Q6: 8x + 4 = 3 (x – 1) + 7
Ans: 8x + 4 = 3 (x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x – 3x = 4 – 4
⇒ 5x = 0
⇒ x = 0
Putting the value of x in RHS and LHS, we get, (8×0) + 4 = 3 (0 – 1) + 7
⇒ 0 + 4 = 0 – 3 + 7
⇒ 4 = 4
⇒ LHS = RHS
Q7: x = 4/5 (x + 10)
Ans: x = 4/5 (x + 10)
⇒ x = 4x/5 + 40/5
⇒ x – (4x/5) = 8
⇒ (5x – 4x)/5 = 8
⇒ x = 8 × 5
⇒ x = 40
Putting the value of x in RHS and LHS, we get,
40 = 4/5 (40 + 10)
⇒ 40 = 4/5 × 50
⇒ 40 = 200/5
⇒ 40 = 40
⇒ LHS = RHS
Q8: 2x/3 + 1 = 7x/15 + 3
Ans: 2x/3 + 1 = 7x/15 + 3
⇒ 2x/3 – 7x/15 = 3 – 1
⇒ (10x – 7x)/15 = 2
⇒ 3x = 2 × 15
⇒ 3x = 30
⇒ x = 30/3
⇒ x = 10
Putting the value of x in RHS and LHS, we get,
Q9: 2y + 5/3 = 26/3 – y
Ans: 2y + 5/3 = 26/3 – y
⇒ 2y + y = 26/3 – 5/3
⇒ 3y = (26 – 5)/3
⇒ 3y = 21/3
⇒ 3y = 7
⇒ y = 7/3
Putting the value of y in RHS and LHS, we get,
⇒ (2 × 7/3) + 5/3 = 26/3 – 7/3
⇒ 14/3 + 5/3 = 26/3 – 7/3
⇒ (14 + 5)/3 = (26 – 7)/3
⇒ 19/3 = 19/3
⇒ LHS = RHS
Q10: 3m = 5m – 8/5
Ans: 3m = 5m – 8/5
⇒ 5m – 3m = 8/5
⇒ 2m = 8/5
⇒ 2m × 5 = 8
⇒ 10m = 8
⇒ m = 8/10
⇒ m = 4/5
Putting the value of m in RHS and LHS, we get,
⇒ 3 × (4/5) = (5 × 4/5) – 8/5
⇒ 12/5 = 4 – (8/5)
⇒ 12/5 = (20 – 8)/5
⇒ 12/5 = 12/5
⇒ LHS = RHS
Exercise: 2.2
Solve the following linear equations.
Q1: x/2 – 1/5 = x/3 +1/4
Ans: x/2 – 1/5 = x/3 +1/4
⇒ x/2 – x/3 =1/4+ 1/5
⇒ (3x – 2x)/6 = (5 + 4)/20
⇒ 3x – 2x = 9/20 × 6
⇒ x = 54/20
⇒ x = 27/10
Q2: n/2 – 3n/4 + 5n/6 = 21
Ans: n/2 – 3n/4 + 5n/6 = 21
⇒ (6n – 9n + 10n)/12 = 21
⇒ 7n/12 = 21
⇒ 7n = 21 × 12
⇒ n = 252/7
⇒ n = 36
Q3: x + 7 – 8x/3 = 17/6 – 5x/2
Ans: x + 7 – 8x/3 = 17/6 – 5x/2
⇒ x – 8x/3 + 5x/2 = 17/6 – 7
⇒ (6x – 16x + 15x)/6 = (17 – 42)/6
⇒ 5x/6 = – 25/6
⇒ 5x = – 25
⇒ x = – 5
Q4: (x – 5)/3 = (x – 3)/5
Ans: (x – 5)/3 = (x – 3)/5
⇒ 5(x-5) = 3(x-3)
⇒ 5x-25 = 3x-9
⇒ 5x – 3x = -9+25
⇒ 2x = 16
⇒ x = 8
Q5: (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
Ans: (3t – 2)/4 – (2t + 3)/3 = 2/3 – t
⇒ ((3t – 2)/4) × 12 – ((2t + 3)/3) × 12
⇒ (3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t
⇒ 9t – 6 – 8t – 12 = 8 – 12t
⇒ t – 18 = 8 – 12t
⇒ t + 12t = 8 + 18
⇒ 13t = 26
⇒ t = 2
Q6: m – (m – 1)/2 = 1 – (m – 2)/3
Ans: m – (m – 1)/2 = 1 – (m – 2)/3
⇒ m – m/2 – 1/2 = 1 – (m/3 – 2/3)
⇒ m – m/2 + 1/2 = 1 – m/3 + 2/3
⇒ m – m/2 + m/3 = 1 + 2/3 – 1/2
⇒ m/2 + m/3 = 1/2 + 2/3
⇒ (3m + 2m)/6 = (3 + 4)/6
⇒ 5m/6 = 7/6
⇒ m = 7/6 × 6/5
⇒ m = 7/5
Simplify and solve the following linear equations.
Q7: 3 (t – 3) = 5(2t + 1)
Ans: 3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5
⇒ 3t – 10t = 5 + 9
⇒ -7t = 14
⇒ t = 14/-7
⇒ t = -2
Q8: 15(y – 4) –2(y – 9) + 5(y + 6) = 0
Ans: 15(y – 4) –2(y – 9) + 5(y + 6) = 0
⇒ 15y – 60 -2y + 18 + 5y + 30 = 0
⇒ 15y – 2y + 5y = 60 – 18 – 30
⇒ 18y = 12
⇒ y = 12/18
⇒ y = 2/3
Q9: 3 (5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Ans: 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
⇒ 15z – 18z – 32z = -52 – 17 + 21 – 22
⇒ -35z = -70
⇒ z = -70/-35
⇒ z = 2
Q10: 0.25(4f – 3) = 0.05(10f – 9)
Ans: 0.25(4f – 3) = 0.05(10f – 9)
⇒ f – 0.75 = 0.5f – 0.45
⇒ f – 0.5f = -0.45 + 0.75
⇒ 0.5f = 0.30
⇒ f = 0.30/0.5
⇒ f = 3/5
⇒ f = 0.6
The document NCERT Solutions for Class 8 Maths - Linear Equations in One Variable (Exercise 2.1) is a part of the Class 8 Course Mathematics (Maths) Class 8.
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FAQs on NCERT Solutions for Class 8 Maths - Linear Equations in One Variable (Exercise 2.1)
| 1. What are linear equations in one variable? |  |
| 2. How do you solve a linear equation in one variable? | |
Ans. To solve a linear equation in one variable, you need to isolate the variable on one side of the equation. This typically involves performing operations such as addition, subtraction, multiplication, or division on both sides of the equation until you get the variable by itself. For example, for the equation 2x + 3 = 7, you would subtract 3 from both sides to get 2x = 4, and then divide by 2 to find x = 2.
| 3. What is the importance of solutions to linear equations in real life? | |
Ans. Solutions to linear equations in real life are important because they can model various situations such as budgeting, calculating distances, and predicting profits. For instance, if a business wants to determine how many products to sell to break even, it can use a linear equation to find the necessary sales volume based on costs and revenues.
| 4. Can a linear equation have more than one solution? | |
Ans. No, a linear equation in one variable can only have one solution. This is because a straight line intersects the x-axis at only one point, which corresponds to the single value of the variable that satisfies the equation. However, if the equation is inconsistent (like 2x + 3 = 2x + 5), it has no solution.
| 5. What are some common mistakes to avoid when solving linear equations? | |
Ans. Common mistakes when solving linear equations include forgetting to apply the same operation to both sides of the equation, miscalculating arithmetic operations, and not properly isolating the variable. It is important to carefully check each step to ensure accuracy and avoid such errors.
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