RD Sharma Class 7 ex 8.1 Solutions Chapter 8 Linear Equations in One Variable

Chapter 8: Linear Equations in One Variable Exercise – 8.1

In this chapter, we provide RD Sharma Class 7 ex 8.1 Solutions Chapter 8 Linear Equations in One Variable for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 8.1 Solutions Chapter 8 Linear Equations in One Variable Maths pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 8
Chapter NameLinear Equations in One Variable
ExerciseEx 8.1

RD Sharma Solutions for Class 7 Chapter 8 Linear Equations in One Variable Ex 8.1 Download PDF

Question: 1

Verify by substitution that:

(i) x = 4 is the root of 3x – 5 = 7

(ii) x = 3 is the root of 5 + 3x = 14

(iii) x = 2 is the root of 3x – 2 = 8x – 12

(iv) x = 4 is the root of 3x/2 = 6

(v) y = 2 is the root of y – 3 = 2y – 5

(vi) x = 8 is the root of (1/2)x + 7 = 11

Solution:

(i) x = 4 is the root of 3x – 5 = 7.

Now, substituting x = 4 in place of ‘x’ in the given equation 3x – 5 = 7,

3(4) – 5 = 7

12 – 5 = 7

7 = 7

Since, LHS = RHS

Hence, x = 4 is the root of 3x – 5 = 7.

(ii). x = 3 is the root of 5 + 3x = 14.

Now, substituting x = 3 in place of ‘x’ in the given equation 5 + 3x = 14,

5 + 3(3) = 14

5 + 9 = 14

14 = 14

Since, LHS = RHS

Hence, x = 3 is the root of 5 + 3x = 14.

(iii). x = 2 is the root of 3x – 2 = 8x – 12.

Now, substituting x = 2 in place of ‘x’ in the given equation 3x –2 = 8x – 12,

3(2) – 2 = 8(2) – 12

6 – 2 = 16 – 12

4 = 4

Since, LHS = RHS

Hence, x = 2 is the root of 3x – 2 = 8x – 12.

(iv) x = 4 is the root of 3x/2 = 6.

Now, substituting x = 4 in place of ‘x’ in the given equation 3x/2 = 6,

(3× 4)/2 = 6

12/2 = 6

6 = 6

Since, LHS = RHS

Hence, x = 4 is the root of 3x/2 = 6.

(v). y = 2 is the root of y – 3 = 2y – 5.

Now, substituting y = 2 in place of ‘y’ in the given equation y – 3 = 2y – 5,

2 – 3 = 2(2) – 5

-1 = 4 – 5

-1 = -1

Since, LHS = RHS

Hence, y = 2 is the root of y – 3 = 2y – 5.

(vi). x = 8 is the root of 12x + 7 = 11.

Now, substituting x = 8 in place of ‘x’ in the given equation 12x + 7 = 11,

12(8) + 7 =11

4 + 7 = 11

11 = 11

Since, LHS = RHS

Hence, x = 8 is the root of 12x + 7 = 11.

Question: 2

Solve each of the following equations by trial and error method:

(i) x + 3 = 12

(ii) x – 7 = 10

(iii) 4x = 28

(iv) x/2 + 7 = 11

(v) 2x + 4 = 3x

(vi) x/4 = 12

(vii) 15/x = 3

(viii) x/18 = 20

Solution:

(i) x + 3 = 12

Here, LHS = x + 3 and RHS = 12

xLHSRHSIs LHS = RHS
11 + 3 = 412No
22 + 3 = 512No
33 + 3 = 612No
44 + 3 = 712No
55 + 3 = 812No
66 + 3 = 912No
77 + 3 = 1012No
88 + 3 = 1112No
99 + 3 = 1212Yes

Therefore, if x = 9, LHS = RHS.

Hence, x = 9 is the solution to this equation.

(ii) x – 7 = 10

Here, LHS = x – 7 and RHS = 10.

xLHSRHSIs LHS = RHS
99 – 7 = 210No
1010 – 7 = 310No
1111 – 7 = 410No
1212 – 7 = 510No
1313 – 7 = 610No
1414 – 7 = 710No
1515 – 7 = 810No
1616 – 7 = 910No
1717 – 7 = 1010Yes

Therefore, if x = 17, LHS = RHS.

Hence, x = 17 is the solution to this equation.

(iii) 4x = 28

Here, LHS = 4x and RHS = 28.

xLHSRHSIs LHS = RHS
14 x 1 = 428No
24 x 2 = 828No
34 x 3 = 1228No
44 x 4 = 1628No
54 x 5 = 2028No
64 x 6 = 2428No
74 x 7 = 2828Yes

Therefore, if x = 7, LHS = RHS

Hence, x = 7 is the solution to this equation.

(iv) x/2 + 7 = 11

Here, LHS = x/2 + 7 and RHS = 11.

Since RHS is a natural number, x/2 must also be a natural number, so we must substitute values of x that are multiples of 2.

xLHSRHSIs LHS = RHS
22/2 + 7 = 811No
44/2 + 7 = 911No
66/2 + 7 = 1011No
88/2 + 7 = 1111Yes

Therefore, if x = 8, LHS = RHS.

Hence, x = 8 is the solution to this equation.

(v) 2x + 4 = 3x

Here, LHS = 2x + 4 and RHS = 3x.

xLHSRHSIs LHS = RHS
12(1) + 4 = 63(1) = 3No
22(2) + 4 = 83(2) = 6No
32(3) + 4 =103(3) = 9No
42(4) + 5 = 12 3(4) = 12Yes

Therefore, if x = 4, LHS = RHS.

Hence, x = 4 is the solution to this equation.

(vi) x/4 = 12

Here, LHS = x/4 and RHS = 12.

Since RHS is a natural number, x/4 must also be a natural number, so we must substitute values of x that are multiples of 4.

XLHSRHSIs LHS = RHS
1616/4 = 412No
2020/4 = 512No
2424/4 = 612No
2828/4 = 712No
3232/4 = 812No
3636/4 = 912No
4040/4 = 1012No
4444/4 = 1112No
4848/4 = 1212Yes

Therefore, if x = 48, LHS = RHS.

Hence, x = 48 is the solution to this equation.

(vii) 15x = 3

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