Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > RS Aggarwal Solutions: Exercise 8A - Linear Equations
RS Aggarwal Solutions: Exercise 8A - Linear Equations | Mathematics (Maths) Class 8 PDF Download
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Page 1
Question 1.
Solve:
8x + 3 = 27 + 2x
Answer:
8x + 3 = 27 + 2x
By transposition,
? 8x – 2x = 27 - 3
? 6x = 24
? x = 4
Question 2.
Solve:
5x + 7 = 2x - 8
Answer:
5x + 7 = 2x - 8
By transposition,
? 5x-2x = - 8 - 7
? 3x = - 15
? x = - 5
Question 3.
Solve:
2z - 1 = 14 - z
Answer:
2z - 1 = 14 - z
By transposition,
? 2z + z = 14 + 1
? 3z = 15
Dividing by 3, on both the sides we get,
Page 2
Question 1.
Solve:
8x + 3 = 27 + 2x
Answer:
8x + 3 = 27 + 2x
By transposition,
? 8x – 2x = 27 - 3
? 6x = 24
? x = 4
Question 2.
Solve:
5x + 7 = 2x - 8
Answer:
5x + 7 = 2x - 8
By transposition,
? 5x-2x = - 8 - 7
? 3x = - 15
? x = - 5
Question 3.
Solve:
2z - 1 = 14 - z
Answer:
2z - 1 = 14 - z
By transposition,
? 2z + z = 14 + 1
? 3z = 15
Dividing by 3, on both the sides we get,
? z = 5
Question 4.
Solve:
9x + 5 = 4(x - 2) + 8
Answer:
9x + 5 = 4(x - 2) + 8
By transposition,
? 9x + 5 = 4x - 8 + 8
? 9x - 4x = - 5 + 0
? 5x = - 5
? x = - 1
Question 5.
Solve:
Answer:
By cross multiplication
Taking LCM of 5 and 1 = 5 on LHS
? 2y = - 5 × 4
? y = - 5 × 2 = - 10
?
Page 3
Question 1.
Solve:
8x + 3 = 27 + 2x
Answer:
8x + 3 = 27 + 2x
By transposition,
? 8x – 2x = 27 - 3
? 6x = 24
? x = 4
Question 2.
Solve:
5x + 7 = 2x - 8
Answer:
5x + 7 = 2x - 8
By transposition,
? 5x-2x = - 8 - 7
? 3x = - 15
? x = - 5
Question 3.
Solve:
2z - 1 = 14 - z
Answer:
2z - 1 = 14 - z
By transposition,
? 2z + z = 14 + 1
? 3z = 15
Dividing by 3, on both the sides we get,
? z = 5
Question 4.
Solve:
9x + 5 = 4(x - 2) + 8
Answer:
9x + 5 = 4(x - 2) + 8
By transposition,
? 9x + 5 = 4x - 8 + 8
? 9x - 4x = - 5 + 0
? 5x = - 5
? x = - 1
Question 5.
Solve:
Answer:
By cross multiplication
Taking LCM of 5 and 1 = 5 on LHS
? 2y = - 5 × 4
? y = - 5 × 2 = - 10
?
Question 6.
Solve:
3x + = 2x + 1
Answer:
3x + = 2x + 1
By cross multiplication
Taking LCM of 3 and 1 = 3 on RHS
Question 7.
Solve:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Answer:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Opening the brackets and multiplying, we get,
? 15y - 60 - 2y + 18 + 5y + 30 = 0
? 15y - 2y + 5y - 60 + 18 + 30 = 0
? 18y = 12
Question 8.
Solve:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Answer:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Multiplying we get,
? 15x - 21 - 18x + 22 = 32x - 52 – 17
Solving, we get
(15x - 18x) + (22 - 21) = 32x - (52 + 17)
Page 4
Question 1.
Solve:
8x + 3 = 27 + 2x
Answer:
8x + 3 = 27 + 2x
By transposition,
? 8x – 2x = 27 - 3
? 6x = 24
? x = 4
Question 2.
Solve:
5x + 7 = 2x - 8
Answer:
5x + 7 = 2x - 8
By transposition,
? 5x-2x = - 8 - 7
? 3x = - 15
? x = - 5
Question 3.
Solve:
2z - 1 = 14 - z
Answer:
2z - 1 = 14 - z
By transposition,
? 2z + z = 14 + 1
? 3z = 15
Dividing by 3, on both the sides we get,
? z = 5
Question 4.
Solve:
9x + 5 = 4(x - 2) + 8
Answer:
9x + 5 = 4(x - 2) + 8
By transposition,
? 9x + 5 = 4x - 8 + 8
? 9x - 4x = - 5 + 0
? 5x = - 5
? x = - 1
Question 5.
Solve:
Answer:
By cross multiplication
Taking LCM of 5 and 1 = 5 on LHS
? 2y = - 5 × 4
? y = - 5 × 2 = - 10
?
Question 6.
Solve:
3x + = 2x + 1
Answer:
3x + = 2x + 1
By cross multiplication
Taking LCM of 3 and 1 = 3 on RHS
Question 7.
Solve:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Answer:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Opening the brackets and multiplying, we get,
? 15y - 60 - 2y + 18 + 5y + 30 = 0
? 15y - 2y + 5y - 60 + 18 + 30 = 0
? 18y = 12
Question 8.
Solve:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Answer:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Multiplying we get,
? 15x - 21 - 18x + 22 = 32x - 52 – 17
Solving, we get
(15x - 18x) + (22 - 21) = 32x - (52 + 17)
? - 3x + 1 = 32x - 69
? 35x = 70
? x = 2
Question 9.
Solve:
Answer:
Taking LCM of 2 and 5 = 10 on LHS
By cross multiplication
? 5x - 25 - 2x + 6 = 10/2
? 3x = 5 + 19
Question 10.
Solve:
Answer:
Taking LCM of 3 and 4 = 12 on LHS
and LCM of 3 and 1 = 3 on RHS
Page 5
Question 1.
Solve:
8x + 3 = 27 + 2x
Answer:
8x + 3 = 27 + 2x
By transposition,
? 8x – 2x = 27 - 3
? 6x = 24
? x = 4
Question 2.
Solve:
5x + 7 = 2x - 8
Answer:
5x + 7 = 2x - 8
By transposition,
? 5x-2x = - 8 - 7
? 3x = - 15
? x = - 5
Question 3.
Solve:
2z - 1 = 14 - z
Answer:
2z - 1 = 14 - z
By transposition,
? 2z + z = 14 + 1
? 3z = 15
Dividing by 3, on both the sides we get,
? z = 5
Question 4.
Solve:
9x + 5 = 4(x - 2) + 8
Answer:
9x + 5 = 4(x - 2) + 8
By transposition,
? 9x + 5 = 4x - 8 + 8
? 9x - 4x = - 5 + 0
? 5x = - 5
? x = - 1
Question 5.
Solve:
Answer:
By cross multiplication
Taking LCM of 5 and 1 = 5 on LHS
? 2y = - 5 × 4
? y = - 5 × 2 = - 10
?
Question 6.
Solve:
3x + = 2x + 1
Answer:
3x + = 2x + 1
By cross multiplication
Taking LCM of 3 and 1 = 3 on RHS
Question 7.
Solve:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Answer:
15(y - 4) - 2(y - 9) + 5(y + 6) = 0
Opening the brackets and multiplying, we get,
? 15y - 60 - 2y + 18 + 5y + 30 = 0
? 15y - 2y + 5y - 60 + 18 + 30 = 0
? 18y = 12
Question 8.
Solve:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Answer:
3(5x - 7) - 2(9x - 11) = 4(8x - 13) - 17
Multiplying we get,
? 15x - 21 - 18x + 22 = 32x - 52 – 17
Solving, we get
(15x - 18x) + (22 - 21) = 32x - (52 + 17)
? - 3x + 1 = 32x - 69
? 35x = 70
? x = 2
Question 9.
Solve:
Answer:
Taking LCM of 2 and 5 = 10 on LHS
By cross multiplication
? 5x - 25 - 2x + 6 = 10/2
? 3x = 5 + 19
Question 10.
Solve:
Answer:
Taking LCM of 3 and 4 = 12 on LHS
and LCM of 3 and 1 = 3 on RHS
By cross multiplication
? 9t - 6 - 8t - 12 = 4(2 - 3t)
? 9t - 6 - 8t - 12 = 8 - 12t
? t - 18 = 8 - 12t
? t + 12t = 8 + 18
Question 11.
Solve:
Answer:
Taking LCM of 5 and 2 = 10 on LHS and LCM of 3 and 1 = 3 on RHS
By cross multiplication
? 3(4x + 14 - 15x - 55) = 10(2x - 7)
? 3( - 11x - 41) = 20x - 70
? - 33x - 20x = 123 - 70
Question 12.
Solve:
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FAQs on RS Aggarwal Solutions: Exercise 8A - Linear Equations - Mathematics (Maths) Class 8
| 1. What are linear equations? |  |
Ans. Linear equations are equations in which the highest power of the variable is 1. They can be expressed in the form of ax + b = 0, where a and b are constants and x is the variable.
| 2. How do you solve linear equations? | |
Ans. To solve linear equations, you can use the method of balancing both sides of the equation by performing the same operation on both sides. The goal is to isolate the variable on one side of the equation.
| 3. Can linear equations have more than one solution? | |
Ans. Yes, linear equations can have more than one solution. For example, if the equation is 2x + 4 = 2(x + 2), both sides of the equation are equal for any value of x, so the equation has infinite solutions.
| 4. What is the importance of linear equations in real life? | |
Ans. Linear equations are used in various real-life scenarios, such as calculating distances, determining rates of change, and solving problems involving proportions. They are essential in fields like physics, engineering, economics, and finance.
| 5. Are linear equations only applicable to straight lines? | |
Ans. No, linear equations are not limited to straight lines. While linear equations can represent straight lines on a graph, they can also represent other relationships, such as proportional relationships between variables. The term "linear" refers to the power of the variable, not the shape of the graph.
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Nov 21, 2024 Last updated |
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