CBSE Class 8 > Class 8 Notes > Mathematics (Maths) > RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable
RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable
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Page 1
Page No – 9.11
Solve each of the following equations and also check your results in
each case:
1.
( ???? +?? )
?? = 3x – 10
Solution:
( 2x+5)
3
= 3x – 10
Let us simplify,
( 2x+5)
3
– 3x = – 10
By taking LCM
( 2x + 5 – 9x)
3
= -10
( -7x+5)
3
= -10
By using cross-multiplication we get,
-7x + 5 = -30
-7x = -30 – 5
-7x = -35
x =
-35
-7
= 5
Let us verify the given equation now,
( 2x+5)
3
= 3x – 10
Page 2
Page No – 9.11
Solve each of the following equations and also check your results in
each case:
1.
( ???? +?? )
?? = 3x – 10
Solution:
( 2x+5)
3
= 3x – 10
Let us simplify,
( 2x+5)
3
– 3x = – 10
By taking LCM
( 2x + 5 – 9x)
3
= -10
( -7x+5)
3
= -10
By using cross-multiplication we get,
-7x + 5 = -30
-7x = -30 – 5
-7x = -35
x =
-35
-7
= 5
Let us verify the given equation now,
( 2x+5)
3
= 3x – 10
By substituting the value of ‘x’ we get,
( 2×5+5)
3
= 3(5) – 10
( 10+5)
3
= 15 – 10
15
3
= 5
5 = 5
Hence, the given equation is verified
2.
( ?? -?? )
?? =
( ?? -?? )
??
Solution:
( a-8)
3
=
( a-3)
2
By using cross-multiplication we get,
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
-a = 7
a = -7
Let us verify the given equation now,
( a-8)
3
=
( a-3)
2
By substituting the value of ‘a’ we get,
( -7 – 8)
3
=
( -7 – 3)
2
Page 3
Page No – 9.11
Solve each of the following equations and also check your results in
each case:
1.
( ???? +?? )
?? = 3x – 10
Solution:
( 2x+5)
3
= 3x – 10
Let us simplify,
( 2x+5)
3
– 3x = – 10
By taking LCM
( 2x + 5 – 9x)
3
= -10
( -7x+5)
3
= -10
By using cross-multiplication we get,
-7x + 5 = -30
-7x = -30 – 5
-7x = -35
x =
-35
-7
= 5
Let us verify the given equation now,
( 2x+5)
3
= 3x – 10
By substituting the value of ‘x’ we get,
( 2×5+5)
3
= 3(5) – 10
( 10+5)
3
= 15 – 10
15
3
= 5
5 = 5
Hence, the given equation is verified
2.
( ?? -?? )
?? =
( ?? -?? )
??
Solution:
( a-8)
3
=
( a-3)
2
By using cross-multiplication we get,
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
-a = 7
a = -7
Let us verify the given equation now,
( a-8)
3
=
( a-3)
2
By substituting the value of ‘a’ we get,
( -7 – 8)
3
=
( -7 – 3)
2
-15
3
=
-10
2
-5 = -5
Hence, the given equation is verified
3.
( ???? + ?? )
?? =
( ???? – ?? )
????
Solution:
( 7y+2)
5
=
( 6y – 5)
11
By using cross-multiplication we get,
(7y + 2)11 = (6y – 5)5
77y + 22 = 30y – 25
77y – 30y = -25 – 22
47y = -47
y =
-47
47
y = -1
Let us verify the given equation now,
( 7y+2)
5
=
( 6y – 5)
11
By substituting the value of ‘y’ we get,
( 7( -1) + 2)
5
=
( 6( -1) – 5)
11
( -7+2)
5
=
( -6 – 5)
11
-
5
5
= -
11
11
Page 4
Page No – 9.11
Solve each of the following equations and also check your results in
each case:
1.
( ???? +?? )
?? = 3x – 10
Solution:
( 2x+5)
3
= 3x – 10
Let us simplify,
( 2x+5)
3
– 3x = – 10
By taking LCM
( 2x + 5 – 9x)
3
= -10
( -7x+5)
3
= -10
By using cross-multiplication we get,
-7x + 5 = -30
-7x = -30 – 5
-7x = -35
x =
-35
-7
= 5
Let us verify the given equation now,
( 2x+5)
3
= 3x – 10
By substituting the value of ‘x’ we get,
( 2×5+5)
3
= 3(5) – 10
( 10+5)
3
= 15 – 10
15
3
= 5
5 = 5
Hence, the given equation is verified
2.
( ?? -?? )
?? =
( ?? -?? )
??
Solution:
( a-8)
3
=
( a-3)
2
By using cross-multiplication we get,
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
-a = 7
a = -7
Let us verify the given equation now,
( a-8)
3
=
( a-3)
2
By substituting the value of ‘a’ we get,
( -7 – 8)
3
=
( -7 – 3)
2
-15
3
=
-10
2
-5 = -5
Hence, the given equation is verified
3.
( ???? + ?? )
?? =
( ???? – ?? )
????
Solution:
( 7y+2)
5
=
( 6y – 5)
11
By using cross-multiplication we get,
(7y + 2)11 = (6y – 5)5
77y + 22 = 30y – 25
77y – 30y = -25 – 22
47y = -47
y =
-47
47
y = -1
Let us verify the given equation now,
( 7y+2)
5
=
( 6y – 5)
11
By substituting the value of ‘y’ we get,
( 7( -1) + 2)
5
=
( 6( -1) – 5)
11
( -7+2)
5
=
( -6 – 5)
11
-
5
5
= -
11
11
-1 = -1
Hence, the given equation is verified
4. x – 2x + 2 –
???? ?? x + 5 = 3 –
?? ?? x
Solution:
x – 2x + 2 –
16
3
x + 5 = 3 –
7
2
x
Let us rearrange the equation
x – 2x –
16x
3
+
7x
2
= 3 – 2 – 5
By taking LCM for 2 and 3 which is 6
( 6x – 12x – 32x + 21x)
6
= -4
-
17x
6
= -4
By cross-multiplying
-17x = -4 × 6
-17x = -24
x =
-24
-17
x =
24
17
Let us verify the given equation now,
x – 2x + 2 –
16
3
x + 5 = 3 –
7
2
x
By substituting the value of ‘x’ we get,
24
17
- 2 (
24
17
) + 2 - (
16
3
)(
24
17
) + 5 = 3 - (
7
2
)(
24
17
)
Page 5
Page No – 9.11
Solve each of the following equations and also check your results in
each case:
1.
( ???? +?? )
?? = 3x – 10
Solution:
( 2x+5)
3
= 3x – 10
Let us simplify,
( 2x+5)
3
– 3x = – 10
By taking LCM
( 2x + 5 – 9x)
3
= -10
( -7x+5)
3
= -10
By using cross-multiplication we get,
-7x + 5 = -30
-7x = -30 – 5
-7x = -35
x =
-35
-7
= 5
Let us verify the given equation now,
( 2x+5)
3
= 3x – 10
By substituting the value of ‘x’ we get,
( 2×5+5)
3
= 3(5) – 10
( 10+5)
3
= 15 – 10
15
3
= 5
5 = 5
Hence, the given equation is verified
2.
( ?? -?? )
?? =
( ?? -?? )
??
Solution:
( a-8)
3
=
( a-3)
2
By using cross-multiplication we get,
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
-a = 7
a = -7
Let us verify the given equation now,
( a-8)
3
=
( a-3)
2
By substituting the value of ‘a’ we get,
( -7 – 8)
3
=
( -7 – 3)
2
-15
3
=
-10
2
-5 = -5
Hence, the given equation is verified
3.
( ???? + ?? )
?? =
( ???? – ?? )
????
Solution:
( 7y+2)
5
=
( 6y – 5)
11
By using cross-multiplication we get,
(7y + 2)11 = (6y – 5)5
77y + 22 = 30y – 25
77y – 30y = -25 – 22
47y = -47
y =
-47
47
y = -1
Let us verify the given equation now,
( 7y+2)
5
=
( 6y – 5)
11
By substituting the value of ‘y’ we get,
( 7( -1) + 2)
5
=
( 6( -1) – 5)
11
( -7+2)
5
=
( -6 – 5)
11
-
5
5
= -
11
11
-1 = -1
Hence, the given equation is verified
4. x – 2x + 2 –
???? ?? x + 5 = 3 –
?? ?? x
Solution:
x – 2x + 2 –
16
3
x + 5 = 3 –
7
2
x
Let us rearrange the equation
x – 2x –
16x
3
+
7x
2
= 3 – 2 – 5
By taking LCM for 2 and 3 which is 6
( 6x – 12x – 32x + 21x)
6
= -4
-
17x
6
= -4
By cross-multiplying
-17x = -4 × 6
-17x = -24
x =
-24
-17
x =
24
17
Let us verify the given equation now,
x – 2x + 2 –
16
3
x + 5 = 3 –
7
2
x
By substituting the value of ‘x’ we get,
24
17
- 2 (
24
17
) + 2 - (
16
3
)(
24
17
) + 5 = 3 - (
7
2
)(
24
17
)
24
17
-
48
17
+ 2 –
384
51
+ 5 = 3 -
168
34
By taking 51 and 34 as the LCM we get,
( 72 – 144 + 102 – 384 + 255)
51
=
( 102 – 168)
34
-
99
51
= -
66
34
-
33
17
= -
33
17
Hence, the given equation is verified
5.
?? ?? ?? + ???? - ?? = ???? +
?? ??
Solution:
1
2x
+ 7x - 6 = 7x +
1
4
Let us rearrange the equation
1
2x
+ 7x - 7x =
1
4
+ 6 (by taking LCM)
1
2
x =
( 1+ 24)
4
1
2
x =
25
4
By cross-multiplying
4x = 25 × 2
4x = 50
x =
50
4
x =
25
2
Let us verify the given equation now,
Read MoreFAQs on RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable
| 1. How do I solve linear equations with variables on both sides? |  |
Ans. Move all variable terms to one side using inverse operations, then isolate the variable by performing the same operation on both sides. For example, in 3x + 5 = x + 13, subtract x from both sides to get 2x + 5 = 13, then solve. This balancing method keeps equations equal throughout the solution process.
| 2. What's the difference between an equation and an expression in Class 8 maths? | |
Ans. An expression is a mathematical phrase with numbers and variables (like 2x + 3), while an equation states that two expressions are equal using an equals sign (like 2x + 3 = 11). Equations can be solved to find variable values; expressions can only be simplified. Understanding this distinction helps students correctly approach linear equation problems.
| 3. Why do we get wrong answers when solving linear equations if we don't apply operations to both sides? | |
Ans. An equation represents a balanced scale-whatever operation students perform on one side must happen on the other to maintain equality. Skipping this step breaks the balance and produces incorrect solutions. For instance, adding 5 only to the left side in x - 5 = 10 violates the fundamental principle that both sides must remain equal.
| 4. How do I check if my solution to a linear equation is correct? | |
Ans. Substitute the solution back into the original equation and verify both sides equal the same value. If x = 4 solves 2x + 3 = 11, then 2(4) + 3 should equal 11, which it does. This verification method confirms accuracy and is essential for avoiding careless calculation mistakes in CBSE examinations.
| 5. What common mistakes do students make when solving one-variable linear equations? | |
Ans. Common errors include forgetting to apply operations to both sides, making sign errors when moving terms, and incorrectly distributing negative signs across brackets. Students also sometimes divide by coefficients inconsistently. Using EduRev's worksheets and MCQ tests helps identify these pitfalls early and reinforces correct algebraic techniques through repeated practice.
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Document Description: RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable for Class 8 2026 is part of Mathematics (Maths) Class 8 preparation. The notes and questions for RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable have been prepared according to the Class 8 exam syllabus. Information about RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable covers topics like and RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable Example, for Class 8 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable.
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