Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > RD Sharma Solutions (Ex 2.1): Linear Equation in One Variable
RD Sharma Solutions (Ex 2.1): Linear Equation in One Variable | Mathematics (Maths) Class 8 PDF Download
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Page 1
LINEAR EQUATION IN ONE VARIABLE
Page No – 9.5
Solve each of the following equations and also verify your solution:
1. 9
?? ?? = y – ?? ?? ??
Solution:
We have,
9
1
4
= y - 1
1
3
37
4
= y –
4
3
Upon solving we get,
y =
37
4
+
4
3
By taking LCM for 4 and 3 is 12
y =
( 37×3)
12
+
( 4×4)
12
=
111
12
+
16
12
=
( 111+16)
12
=
127
12
? y =
127
12
Page 2
LINEAR EQUATION IN ONE VARIABLE
Page No – 9.5
Solve each of the following equations and also verify your solution:
1. 9
?? ?? = y – ?? ?? ??
Solution:
We have,
9
1
4
= y - 1
1
3
37
4
= y –
4
3
Upon solving we get,
y =
37
4
+
4
3
By taking LCM for 4 and 3 is 12
y =
( 37×3)
12
+
( 4×4)
12
=
111
12
+
16
12
=
( 111+16)
12
=
127
12
? y =
127
12
2.
???? ?? +
?? ?? = 1
Solution:
We have,
5x
3
+
2
5
= 1
5x
3
= 1 -
2
5
(by taking LCM)
=
( 5-2)
5
By using cross-multiplication we get,
5x
3
=
3
5
5x =
( 3×3)
5
x =
9
( 5×5)
=
9
25
? x =
9
25
3.
?? ?? +
?? ?? +
?? ?? = 13
Solution:
We have,
x
2
+
x
3
+
x
4
= 13
let us take LCM for 2, 3 and 4 which is 12
( x×6)
12
+
( x×4)
12
+
( x×3)
12
= 13
Page 3
LINEAR EQUATION IN ONE VARIABLE
Page No – 9.5
Solve each of the following equations and also verify your solution:
1. 9
?? ?? = y – ?? ?? ??
Solution:
We have,
9
1
4
= y - 1
1
3
37
4
= y –
4
3
Upon solving we get,
y =
37
4
+
4
3
By taking LCM for 4 and 3 is 12
y =
( 37×3)
12
+
( 4×4)
12
=
111
12
+
16
12
=
( 111+16)
12
=
127
12
? y =
127
12
2.
???? ?? +
?? ?? = 1
Solution:
We have,
5x
3
+
2
5
= 1
5x
3
= 1 -
2
5
(by taking LCM)
=
( 5-2)
5
By using cross-multiplication we get,
5x
3
=
3
5
5x =
( 3×3)
5
x =
9
( 5×5)
=
9
25
? x =
9
25
3.
?? ?? +
?? ?? +
?? ?? = 13
Solution:
We have,
x
2
+
x
3
+
x
4
= 13
let us take LCM for 2, 3 and 4 which is 12
( x×6)
12
+
( x×4)
12
+
( x×3)
12
= 13
6x
12
+
4x
12
+
3x
12
= 13
( 6x+4x+3x)
12
= 13
13x
12
= 13
By using cross-multiplication we get,
13x = 12 × 13
x =
156
13
= 12
? x = 12
4.
?? ?? +
?? ?? =
?? ??
Solution:
We have,
x
2
+
x
8
=
1
8
let us take LCM for 2 and 8 which is 8
( x×4)
8
+
( x×1)
8
=
1
8
4x
8
+
x
8
=
1
8
5x
8
=
1
8
By using cross-multiplication we get,
5x =
8
8
5x = 1
Page 4
LINEAR EQUATION IN ONE VARIABLE
Page No – 9.5
Solve each of the following equations and also verify your solution:
1. 9
?? ?? = y – ?? ?? ??
Solution:
We have,
9
1
4
= y - 1
1
3
37
4
= y –
4
3
Upon solving we get,
y =
37
4
+
4
3
By taking LCM for 4 and 3 is 12
y =
( 37×3)
12
+
( 4×4)
12
=
111
12
+
16
12
=
( 111+16)
12
=
127
12
? y =
127
12
2.
???? ?? +
?? ?? = 1
Solution:
We have,
5x
3
+
2
5
= 1
5x
3
= 1 -
2
5
(by taking LCM)
=
( 5-2)
5
By using cross-multiplication we get,
5x
3
=
3
5
5x =
( 3×3)
5
x =
9
( 5×5)
=
9
25
? x =
9
25
3.
?? ?? +
?? ?? +
?? ?? = 13
Solution:
We have,
x
2
+
x
3
+
x
4
= 13
let us take LCM for 2, 3 and 4 which is 12
( x×6)
12
+
( x×4)
12
+
( x×3)
12
= 13
6x
12
+
4x
12
+
3x
12
= 13
( 6x+4x+3x)
12
= 13
13x
12
= 13
By using cross-multiplication we get,
13x = 12 × 13
x =
156
13
= 12
? x = 12
4.
?? ?? +
?? ?? =
?? ??
Solution:
We have,
x
2
+
x
8
=
1
8
let us take LCM for 2 and 8 which is 8
( x×4)
8
+
( x×1)
8
=
1
8
4x
8
+
x
8
=
1
8
5x
8
=
1
8
By using cross-multiplication we get,
5x =
8
8
5x = 1
x =
1
5
? x =
1
5
5.
???? ?? -
???? ?? =
?? ????
Solution:
We have,
2x
3
-
3x
8
=
7
12
By taking LCM for 3 and 8 is 24
( 2x×8)
24
-
( 3x×3)
24
=
7
12
16x
24
–
9x
24
=
7
12
( 16x-9x)
24
=
7
12
7x
24
=
7
12
By using cross-multiplication we get,
7x × 12 = 7 × 24
x =
( 7×24)
( 7×12)
=
24
12
= 2
? x = 2
6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0
Page 5
LINEAR EQUATION IN ONE VARIABLE
Page No – 9.5
Solve each of the following equations and also verify your solution:
1. 9
?? ?? = y – ?? ?? ??
Solution:
We have,
9
1
4
= y - 1
1
3
37
4
= y –
4
3
Upon solving we get,
y =
37
4
+
4
3
By taking LCM for 4 and 3 is 12
y =
( 37×3)
12
+
( 4×4)
12
=
111
12
+
16
12
=
( 111+16)
12
=
127
12
? y =
127
12
2.
???? ?? +
?? ?? = 1
Solution:
We have,
5x
3
+
2
5
= 1
5x
3
= 1 -
2
5
(by taking LCM)
=
( 5-2)
5
By using cross-multiplication we get,
5x
3
=
3
5
5x =
( 3×3)
5
x =
9
( 5×5)
=
9
25
? x =
9
25
3.
?? ?? +
?? ?? +
?? ?? = 13
Solution:
We have,
x
2
+
x
3
+
x
4
= 13
let us take LCM for 2, 3 and 4 which is 12
( x×6)
12
+
( x×4)
12
+
( x×3)
12
= 13
6x
12
+
4x
12
+
3x
12
= 13
( 6x+4x+3x)
12
= 13
13x
12
= 13
By using cross-multiplication we get,
13x = 12 × 13
x =
156
13
= 12
? x = 12
4.
?? ?? +
?? ?? =
?? ??
Solution:
We have,
x
2
+
x
8
=
1
8
let us take LCM for 2 and 8 which is 8
( x×4)
8
+
( x×1)
8
=
1
8
4x
8
+
x
8
=
1
8
5x
8
=
1
8
By using cross-multiplication we get,
5x =
8
8
5x = 1
x =
1
5
? x =
1
5
5.
???? ?? -
???? ?? =
?? ????
Solution:
We have,
2x
3
-
3x
8
=
7
12
By taking LCM for 3 and 8 is 24
( 2x×8)
24
-
( 3x×3)
24
=
7
12
16x
24
–
9x
24
=
7
12
( 16x-9x)
24
=
7
12
7x
24
=
7
12
By using cross-multiplication we get,
7x × 12 = 7 × 24
x =
( 7×24)
( 7×12)
=
24
12
= 2
? x = 2
6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0
Solution:
We have,
(x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0
Upon expansion we get,
x
2
+ 5x + 6 + x
2
– 5x +6 – 2x
2
– 2x =0
-2x + 12 = 0
By dividing the equation using -2 we get,
x – 6 = 0
x = 6
? x = 6
7.
?? ?? –
?? ?? +
?? ?? +
???? ???? =
?? ??
Solution:
We have,
x
2
–
4
5
+
x
5
+
3x
10
=
1
5
upon solving we get,
x
2
+
x
5
+
3x
10
=
1
5
+
4
5
by taking LCM for 2, 5 and 10 which is 10
( x×5)
10
+
( x×2)
10
+
( 3x×1)
10
=
5
5
5x
10
+
2x
10
+
3x
10
= 1
( 5x+2x+3x)
10
= 1
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FAQs on RD Sharma Solutions (Ex 2.1): 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 algebraic equation that can be written in the form ax + b = 0, where a and b are constants and x is the variable. It represents a straight line on the coordinate plane and has only one unknown variable.
| 2. How to solve a linear equation in one variable? | |
Ans. To solve a linear equation in one variable, follow these steps:
1. Simplify the equation by combining like terms.
2. Isolate the variable term on one side of the equation by performing the inverse operation.
3. Simplify further to get the value of the variable.
4. Check the solution by substituting it back into the original equation.
| 3. Can a linear equation in one variable have multiple solutions? | |
Ans. No, a linear equation in one variable can have at most one solution. The solution represents the value of the variable that satisfies the equation. If there are multiple solutions, it means the equation is not linear or has been incorrectly simplified.
| 4. What is the graphical representation of a linear equation in one variable? | |
Ans. The graphical representation of a linear equation in one variable is a straight line on the coordinate plane. The slope-intercept form, y = mx + c, is commonly used to graph linear equations, where m is the slope of the line and c is the y-intercept (the point where the line intersects the y-axis).
| 5. Can a linear equation in one variable have no solution? | |
Ans. Yes, a linear equation in one variable can have no solution. This occurs when the equation is contradictory or inconsistent. For example, the equation 2x + 3 = 2x + 5 has no solution since no value of x can satisfy the equation. In graphical terms, it represents two parallel lines that never intersect.
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Nov 21, 2024 Last updated |
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