Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable
RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable | Mathematics (Maths) Class 8 PDF Download
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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,
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FAQs on RD Sharma Solutions (Ex 2.2): Linear Equation in One Variable - Mathematics (Maths) Class 8
| 1. What is a linear equation in one variable? |  |
Ans. A linear equation in one variable is an equation that can be expressed in the form ax + b = 0, where a and b are constants, and x is the variable. It represents a straight line on a graph and has only one unknown variable.
| 2. How can I 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 can be done by performing various operations like addition, subtraction, multiplication, and division on both sides of the equation. The goal is to simplify the equation until you obtain the value of the variable.
| 3. Can a linear equation in one variable have multiple solutions? | |
Ans. No, a linear equation in one variable typically has only one solution. This is because a linear equation represents a straight line, which intersects the x-axis at only one point. However, there are some special cases where the equation may have no solution or infinite solutions, but these are exceptions rather than the norm.
| 4. What is the importance of linear equations in one variable? | |
Ans. Linear equations in one variable are widely used in various fields, including mathematics, physics, and engineering. They help in modeling real-world situations and solving problems involving unknown quantities. Linear equations also form the basis for more advanced concepts in algebra and provide a fundamental understanding of mathematical relationships.
| 5. Can I represent a linear equation in one variable graphically? | |
Ans. Yes, a linear equation in one variable can be represented graphically. The equation represents a straight line on a graph, where the x-axis represents the variable and the y-axis represents the dependent variable. By plotting the points on the graph, you can visualize the relationship between the variable and the dependent variable, and also determine the solution to the equation by finding the point of intersection with the x-axis.
About this Document
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Dec 31, 2024 Last updated |
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