Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Linear Equations in One Variable (Exercise 8.3)
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Page 1
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Page 2
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
Page 3
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
x = (-50/2)
x = -25
Verification:
Substituting x = -25 in given equation we get
(-25/2) + (3/2) = (-50/5) – 1
(-25 + 3)/2 = -10 – 1
(-22/2) = -11
-11 = -11
Thus LHS = RHS
Hence, verified.
5. (3/4) (x -1) = (x – 3)
Solution:
Given (3/4) (x -1) = (x – 3)
On simplifying the brackets both sides we get,
(3/4) x – (3/4) = (x – 3)
Now transposing (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
(3x – 4x)/4 = (3 – 12)/4
-x/4 = (-9/4)
Multiply both sides by -4 we get
-x/4 × -4 = (-9/4) × -4
x = 9
Verification:
Substituting x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
6. 3 (x – 3) = 5 (2x + 1)
Solution:
Given 3 (x – 3) = 5 (2x + 1)
Page 4
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
x = (-50/2)
x = -25
Verification:
Substituting x = -25 in given equation we get
(-25/2) + (3/2) = (-50/5) – 1
(-25 + 3)/2 = -10 – 1
(-22/2) = -11
-11 = -11
Thus LHS = RHS
Hence, verified.
5. (3/4) (x -1) = (x – 3)
Solution:
Given (3/4) (x -1) = (x – 3)
On simplifying the brackets both sides we get,
(3/4) x – (3/4) = (x – 3)
Now transposing (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
(3x – 4x)/4 = (3 – 12)/4
-x/4 = (-9/4)
Multiply both sides by -4 we get
-x/4 × -4 = (-9/4) × -4
x = 9
Verification:
Substituting x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
6. 3 (x – 3) = 5 (2x + 1)
Solution:
Given 3 (x – 3) = 5 (2x + 1)
On simplifying the brackets we get,
3x – 9 = 10x + 5
Now transposing 10x to LHS and 9 to RHs
3x – 10x = 5 + 9
-7x = 14
Now dividing both sides by -7 we get
-7x/-7 = 14/-7
x = -2
Verification:
Substituting x = -2 in the given equation we get
3 (-2 – 3) = 5 (-4 + 1)
3 (-5) = 5 (-3)
-15 = - 15
Thus LHS = RHS
Hence, verified.
7. 3x – 2 (2x -5) = 2 (x + 3) – 8
Solution:
Given 3x – 2 (2x -5) = 2 (x + 3) – 8
On simplifying the brackets on both sides, we get
3x – 2 × 2x + 2 × 5 = 2 × x + 2 × 3 – 8
3x – 4x + 10 = 2x + 6 – 8
-x + 10 = 2x – 2
Transposing x to RHS and 2 to LHS, we get
10 + 2 = 2x + x
3x = 12
Dividing both sides by 3, we get
3x/3 = 12/3
x = 4
Verification:
Substituting x = 4 on both sides, we get
3(4) – 2{2(4) – 5} = 2(4 + 3) – 8
12 – 2(8 – 5) = 14 – 8
12 – 6 = 6
6 = 6
Thus LHS = RHS
Page 5
Exercise 8.3 Page No: 8.19
Solve each of the following equations. Also, verify the result in each case.
1. 6x + 5 = 2x + 17
Solution:
Given 6x + 5 = 2x + 17
Transposing 2x to LHS and 5 to RHS, we get
6x – 2x = 17 – 5
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
x = 3
Verification:
Substituting x = 3 in the given equation, we get
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Therefore LHS = RHS
Hence, verified.
2. 2 (5x – 3) – 3 (2x – 1) = 9
Solution:
Given 2 (5x – 3) – 3 (2x – 1) = 9
Simplifying the brackets, we get
2 × 5x – 2 × 3 – 3 × 2x + 3 × 1 = 9
10x – 6 - 6x + 3 = 9
10x – 6x – 6 + 3 = 9
4x – 3 = 9
Adding 3 to both sides, we get
4x – 3 + 3 = 9 + 3
4x = 12
Dividing both sides by 4, we get
4x/4 = 12/4
Therefore x = 3.
Verification:
Substituting x = 3 in LHS, we get
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Thus, LHS = RHS
Hence, verified.
3. (x/2) = (x/3) + 1
Solution:
Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.
4. (x/2) + (3/2) = (2x/5) - 1
Solution:
Given (x/2) + (3/2) = (2x/5) - 1
Transposing (2x/5) to LHS and (3/2) to RHS, then we get
(x/2) – (2x/5) = - 1 – (3/2)
(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]
x/10 = -5/2
Multiplying both sides by 10 we get,
x/10 × 10 = (-5/2) × 10
x = (-50/2)
x = -25
Verification:
Substituting x = -25 in given equation we get
(-25/2) + (3/2) = (-50/5) – 1
(-25 + 3)/2 = -10 – 1
(-22/2) = -11
-11 = -11
Thus LHS = RHS
Hence, verified.
5. (3/4) (x -1) = (x – 3)
Solution:
Given (3/4) (x -1) = (x – 3)
On simplifying the brackets both sides we get,
(3/4) x – (3/4) = (x – 3)
Now transposing (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
(3x – 4x)/4 = (3 – 12)/4
-x/4 = (-9/4)
Multiply both sides by -4 we get
-x/4 × -4 = (-9/4) × -4
x = 9
Verification:
Substituting x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
6. 3 (x – 3) = 5 (2x + 1)
Solution:
Given 3 (x – 3) = 5 (2x + 1)
On simplifying the brackets we get,
3x – 9 = 10x + 5
Now transposing 10x to LHS and 9 to RHs
3x – 10x = 5 + 9
-7x = 14
Now dividing both sides by -7 we get
-7x/-7 = 14/-7
x = -2
Verification:
Substituting x = -2 in the given equation we get
3 (-2 – 3) = 5 (-4 + 1)
3 (-5) = 5 (-3)
-15 = - 15
Thus LHS = RHS
Hence, verified.
7. 3x – 2 (2x -5) = 2 (x + 3) – 8
Solution:
Given 3x – 2 (2x -5) = 2 (x + 3) – 8
On simplifying the brackets on both sides, we get
3x – 2 × 2x + 2 × 5 = 2 × x + 2 × 3 – 8
3x – 4x + 10 = 2x + 6 – 8
-x + 10 = 2x – 2
Transposing x to RHS and 2 to LHS, we get
10 + 2 = 2x + x
3x = 12
Dividing both sides by 3, we get
3x/3 = 12/3
x = 4
Verification:
Substituting x = 4 on both sides, we get
3(4) – 2{2(4) – 5} = 2(4 + 3) – 8
12 – 2(8 – 5) = 14 – 8
12 – 6 = 6
6 = 6
Thus LHS = RHS
Hence, verified.
8. x – (x/4) – (1/2) = 3 + (x/4)
Solution:
Given x – (x/4) – (1/2) = 3 + (x/4)
Transposing (x/4) to LHS and (1/2) to RHS
x – (x/4) - (x/4) = 3 + (1/2)
(4x – x – x)/4 = (6 + 1)/2
2x/4 = 7/2
x/2 = 7/2
x = 7
Verification:
Substituting x = 7 in the given equation we get
7 – (7/4) – (1/2) = 3 + (7/4)
(28 – 7 – 2)/4 = (12 + 7)/4
19/4 = 19/4
Thus LHS = RHS
Hence, verified.
9. (6x – 2)/9 + (3x + 5)/18 = (1/3)
Solution:
Given (6x – 2)/9 + (3x + 5)/18 = (1/3)
(6x (2) – 2 (2) + 3x + 5)/18 = (1/3)
(12x – 4 + 3x + 5)/18 = (1/3)
(15x + 1)/ 18 = (1/3)
Multiplying both sides by 18 we get
(15x + 1)/18 × 18 = (1/3) × 18
15x + 1 = 6
Transposing 1 to RHS, we get
= 15x = 6 – 1
= 15x = 5
Dividing both sides by 15, we get
= 15x/15 = 5/15
=x = 1/3
Verification:
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FAQs on Linear Equations in One Variable (Exercise 8.3) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. How do you solve linear equations in one variable? |  |
Ans. To solve a linear equation in one variable, you need to isolate the variable on one side of the equation. You can do this by performing various operations such as addition, subtraction, multiplication, and division on both sides of the equation. The goal is to simplify the equation and solve for the unknown variable.
| 2. What is the importance of solving linear equations in one variable? | |
Ans. Solving linear equations in one variable is important as it helps in finding the value of the unknown variable. It is a fundamental concept in mathematics and has applications in various fields such as physics, engineering, finance, and economics. It allows us to solve real-life problems and make predictions based on mathematical models.
| 3. Can linear equations in one variable have more than one solution? | |
Ans. No, linear equations in one variable can have only one solution. This is because a linear equation represents a straight line on a graph, and the intersection of a line with the x-axis represents the solution to the equation. Therefore, there can be either one unique solution or no solution at all.
| 4. How can I check if a given value is a solution to a linear equation in one variable? | |
Ans. To check if a given value is a solution to a linear equation in one variable, substitute the value into the equation and simplify both sides. If the resulting equation is true, then the given value is a solution to the equation. If the resulting equation is false, then the given value is not a solution.
| 5. What are the different methods to solve linear equations in one variable? | |
Ans. There are various methods to solve linear equations in one variable, including:
- Addition and subtraction method: Add or subtract the same number from both sides of the equation to isolate the variable.
- Multiplication and division method: Multiply or divide both sides of the equation by the same number to isolate the variable.
- Cross-multiplication method: This method is used for solving equations involving fractions. It involves multiplying both sides of the equation by the denominators to eliminate the fractions.
- Substitution method: This method involves substituting the value of one variable from one equation into the other equation to solve for the other variable.
- Graphical method: This method involves plotting the equation on a graph and finding the x-coordinate of the point where the line intersects the x-axis, which represents the solution to the equation.
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