RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 9- Linear Equations in One Variable
RS Aggarwal Class 6 Math Solution Chapter 9- Linear Equations in One Variable Exercise 9B is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 9- Linear Equations in One Variable
Exercise 9B
1. `x + 5 = 12`
Solution:
`x + 5 = 12`
Subtracting 5 from both the sides:
⇒ `x` + `cancel 5` − `cancel 5` = `12 − 5`
⇒ `x = 7`
Verification:
Substituting `x = 7` in the L.H.S.:
L.H.S.= `x + 5`
= `7 + 5`
= 12
R.H.S.= 12
∵ L.H.S. = R.H.S.
Hence, verified.
2. `x + 3 = – 2`
Solution:
`x + 3 = −2`
Subtracting 3 from both the sides:
⇒ x + `cancel 3` − `cancel 3` = −2 − 3
⇒ `x = −5`
Verification:
Substituting `x = −5` in the L.H.S.:
L.H.S = `x + 3`
= −5 + 3
= −2
∵ R.H.S. = −2
Hence, verified.
3. `x – 7 = 6`
Solution:
`x − 7 = 6`
Adding 7 on both the sides:
⇒ x − `cancel 7` + `cancel 7` = 6 + 7
⇒ x = 13
Verification:
Substituting x = 13 in the L.H.S.:
L.H.S.= `x − 7`
= 13 − 7
= 6
R.H.S.= 6
∵ L.H.S. = R.H.S.
Hence, verified.
4. `x – 2 = – 5`
Solution:
`x − 2 = −5`
Adding 2 on both sides:
⇒ x − `cancel 2` + `cancel 2` = −5 + 2
⇒ `x` = `−3`
Verification:
Substituting x = −3 in the L.H.S.:
L.H.S. = `x – 2`
= −3 − 2
= −5
R.H.S.= −5
So, L.H.S. = R.H.S.
Hence, verified.
5. `3x – 5 = 13`
Solution:
`3x − 5 = 13`
⇒ `3x` − `cancel 5` + `cancel 5` = `13 + 5`
[Adding 5 on both the sides]
⇒ `3x = 18`
⇒ `frac\{cancel {3}x}{cancel 3}` = `frac\{cancel 18^6}{cancel 3}`
[Dividing both the sides by 3]
⇒ `x = 6`
Verification:
Substituting x = 6 in the L.H.S.:
L.H.S.= `3x – 5`
= `3(6) − 5`
= `18 − 5`
= `13`
R.H.S.= `13`
∵ L.H.S. = R.H.S.
Hence, verified.
6. `4x + 7 = 15`
Solution:
`4x + 7 = 15`
⇒ `4x` + `cancel 7` − `cancel 7` = 15 − 7
[Subtracting 7 from both the sides]
⇒ `4x = 8`
⇒ `frac\{cancel {4}x}{cancel 4}` = `frac\{cancel 8^2}{cancel 4}`
[Dividing both the sides by 4]
⇒ `x = 2`
Verification:
Substituting `x = 2` in the L.H.S.:
L.H.S.= `4x + 7`
= `4 × 2 + 7`
= `8 + 7`
= `15`
R.H.S.= 15
∵ L.H.S. = R.H.S.
Hence, verified.
7. `frac\{x}{5}` = 12
Solution:
`frac\{x}{5}` = 12
⇒ `frac\{x}{cancel 5}` × `cancel 5` = 12 × 5
[Multiplying both the sides by 5]
⇒ `x = 60`
Verification:
Substituting `x = 60` in the L.H.S.:
L.H.S.= `frac\{x}{5}`
= `frac\{cancel 60^12}{cancel 5}`
= `12`
R.H.S. = `12`
∵ L.H.S. = R.H.S.
Hence, verified.
8. `frac\{3x}{5}` = `15`
Solution:
`frac\{3x}{5}` = `15`
⇒ `frac\{3x}{cancel 5}` × `cancel 5` = 15 × 5
[Multiplying both the sides by 5]
⇒ `3x = 75`
⇒ `frac\{cancel {3}x}{cancel 3}` = `frac\{cancel 75^25}{cancel 3}`
[Dividing both the sides by 3]
⇒ `x = 25`
Verification:
Substituting `x = 25` in the L.H.S.:
L.H.S. = `frac\{3x}{5}`
= `frac{3 × 25}{5}`
= `frac{cancel 75^15}{cancel 5}`
= `15`
R.H.S .= `15`
∵ L.H.S. = R.H.S.
Hence, verified.
9. `5x – 3 = x + 17`
Solution:
`5x − 3 = x + 17`
⇒ `5x − x = 17 + 3`
[Transposing x to the L.H.S. and 3 to the R.H.S.]
⇒ `4x = 20`
⇒ `frac\{cancel {4}x}{cancel 4}` = `frac\{cancel 20^5}{cancel 4}`
[Dividing both the sides by 4]
⇒ `x = 5`
Verification:
Substituting x = 5 on both the sides:
L.H.S. = `5x – 3`
= `5(5) − 3`
= 25 − 3
= 22
R.H.S. = `x + 17`
= 5 + 17
= 22
∵ L.H.S. = R.H.S.
Hence, verified.
10. `2x – frac\{1}{2}` = `3`
Solution:
`2x` − `frac\{1}{2}` = `3`
⇒ 2x −`cancel (frac\{1}{2})` + `cancel (frac\{1}{2})` = 3 + `frac\{1}{2}`
[Adding `frac\{1}{2}` on both the sides]
⇒ 2x = `frac\{6 + 1}{2}`
⇒ 2x = `frac\{7}{2}`
⇒ `frac\{cancel {2}x}{cancel 2}` = `frac\{7}{2 × 2}`
[Dividing both the sides by 2]
⇒ `x` = `frac\{7}{4}`
Verification:
Substituting x = `frac\{7}{4}` in the L.H.S.
L.H.S. = `2x` − `frac\{1}{2}`
= `cancel 2` × `frac\{7}{cancel 4^2}` − `frac\{1}{2}`
= `frac\{7}{2}` − `frac\{1}{2}`
= `frac\{7 − 1}{2}`
= `frac\{cancel 6^3}{cancel 2}`
= 3
R.H.S.= 3
So, L.H.S. = R.H.S.
Hence, verified.
11. `3(x + 6) = 24`
Solution:
`3(x + 6) = 24`
⇒ `3x + 18 = 24`
⇒ `3x` + `cancel 18` − `cancel 18` = `24 − 18`
[Subtracting 18 from both the sides]
⇒ `3x = 6`
⇒ `frac\{cancel {3}x}{cancel 3}` = `frac\{cancel 6^2}{cancel 3}`
[Dividing both the sides by 3]
⇒ `x = 2`
Verification:
Substituting x = 2 in the L.H.S.:
L.H.S.= `3(x + 6)`
= `3(2 + 6)`
= `3 × 8 = 24`
R.H.S.= 24
So, L.H.S. = R.H.S.
Hence, verified.
12. `6x + 5 = 2x + 17`
Solution:
`6x + 5 = 2x + 17`
⇒ `6x − 2x = 17 − 5`
[Transposing 2x to the L.H.S. and 5 to the R.H.S.]
⇒ `4x = 12`
⇒`frac\{cancel {4}x}{cancel 4}` = `frac\{cancel 12^3}{cancel 4}`
[Dividing both the sides by 4]
⇒ `x = 3`
Verification:
Substituting x = 3 on both the sides:
L.H.S. = `6x + 5`
= `6(3) + 5`
= `18 + 5`
= `23`
R.H.S.= ` 2x + 17`
= `2(3) + 17`
= `6 + 17`
= `23`
So, L.H.S. = R.H.S.
Hence, verified.
13. `frac\{x}{4}` – 8 = 1
Solution:
`frac\{x}{4}` − 8 = 1
⇒ `frac\{x}{4}`− `cancel 8` + `cancel 8` = 1 + 8
[Adding 8 on both the sides]
⇒ `frac\{x}{4}` = 9
⇒ `frac\{x}{cancel 4}` × `cancel 4` = `9 × 4`
[Multiplying both the sides by 4]
⇒ `x = 36`
Verification:
Substituting `x = 36` in the L.H.S.:
L.H.S. = `frac\{x}{4}` – 8
= `frac\{cancel 36^9}{cancel 4}` − 8
= 9 − 8 = 1
R.H.S.= 1
So, L.H.S. = R.H.S.
Hence, verified.
14. `frac\{x}{2}` = `frac\{x}{3}` + 1
Solution:
`frac\{x}{2}` = `frac\{x}{3}` + 1
⇒ `frac\{x}{2}` − `frac\{x}{3}` = 1
[Transposing `frac\{x}{3}` to the L.H.S.]
⇒ `frac\{3x − 2x}{6}` = 1
⇒ `frac\{x}{6}` = 1
⇒ `frac\{x}{cancel 2}` × `cancel 6^3` = 1 × 6
[Multiplying both the sides by 6]
⇒ `x = 6`
Verification:
Substituting `x = 6` on both the sides:
L.H.S.= `frac\{x}{2}`
= `frac\{cancel 6^3}{cancel 2}`
= `3`
R.H.S.= `frac\{x}{3}` + 1
= `frac\{cancel 6^2}{cancel 3}` + 1
= `2 + 1 `
= 3
So, L.H.S. = R.H.S.
Hence, verified.
15. `3(x + 2) – 2(x – 1) = 7`
Solution:
`3(x + 2) − 2(x − 1) = 7`
⇒ `3x + 6 − 2x + 2 = 7`
⇒ `x + 8 = 7`
⇒ `x` + `cancel 8` − `cancel 8` = `7 − 8`
[Subtracting 8 from both the sides]
⇒ `x = −1`
Verification:
Substituting `x = −1` in the L.H.S.:
L.H.S. = `3(x + 2) – 2(x – 1)`
= `3(−1 + 2) − 2(−1− 1)`
= `3(1) − 2(− 2)`
= `3 + 4`
= `7`
R.H.S.= `7`
So, L.H.S. = R.H.S.
Hence, verified.
16. `5(x – 1) + 2 (x + 3) + 6 = 0`
Solution:
`5(x – 1) + 2(x + 3) + 6 = 0`
⇒ `5x – 5 + 2x + 6 + 6 = 0`
⇒ `7x + 7 = 0`
⇒ `x + 1 = 0` (Dividing by 7)
⇒ `x = – 1`
Verification:
Substituting x = –1 in the L.H.S.:
L.H.S.= `5(x – 1) + 2 (x + 3) + 6`
= `5( –1 –1) + 2(–1 + 3) + 6`
= `5(– 2) + 2(2) + 6`
= `– 10 + 4 + 6`
= ` – 10 + 10`
= 0
R.H.S.= 0
Hence, verified.
17. `6(1 – 4 x) + 7(2 + 5x) = 53`
Solution:
`6(1 − 4x) + 7(2 + 5x) = 53`
⇒ `6 − 24x + 14 + 35x = 53`
⇒ `11x + 20 = 53`
⇒ `11x` + `cancel 20` − `cancel 20` = `53 − 20`
[Subtracting 20 from both the sides]
⇒ `11x = 33`
⇒ `frac\{cancel {11}x}{cancel 11}`= `frac\{cancel 33^3}{cancel 11}`
[Dividing both the sides by 11]
⇒ `x = 3`
Verification:
Substituting x = 3 in the L.H.S.:
L.H.S.= `6(1 – 4 x) + 7(2 + 5x)`
= `6(1 − 4 × 3) + 7(2 + 5 × 3)`
= `6(1 − 12) + 7(2 + 15)`
= `6(−11) + 7(17)`
= ` − 66 + 119`
= `53`
R.H.S. = `53`
So, L.H.S. = R.H.S.
Hence, verified.
18. `16 (3x – 5) – 10 (4x – 8) = 40`
Solution:
`16(3x − 5) − 10(4x − 8) = 40`
⇒`48x − 80 − 40x + 80 = 40`
⇒ `8x = 40`
⇒ `frac\{cancel {8}x}{cancel 8}` = `frac\{cancel 40^5}{cancel 8}`
[Dividing both the sides by 8]
⇒ `x = 5`
Verification:
Substituting `x = 5` in the L.H.S.:
L.H.S.= `16 (3x – 5) – 10 (4x – 8)`
= `16(3 × 5 − 5) − 10( 4 × 5 − 8)`
= `16(15 − 5) − 10(20 − 8)`
= `16(10) −10(12)`
= `160 − 120`
= `40`
R.H.S.= `40`
So, L.H.S. = R.H.S.
Hence, verified.
19. `3(x + 6) + 2 (x + 3) = 64`
Solution:
`3(x + 6) + 2(x + 3) = 64`
⇒ `3x + 18 + 2x + 6 = 64`
⇒ `5x + 24 = 64`
⇒ `5x` + `cancel 24` − `cancel 24` = `64 − 24`
[Subtracting 24 from both the sides]
⇒ `5x = 40`
⇒ `frac\{cancel {5}x}{cancel 5}` = `frac\{cancel 40^8}{cancel 5}`
[Dividing both the sides by 5]
⇒ `x = 8`
Verification:
Substituting `x = 8` in the L.H.S.:
L.H.S. = `3(x + 6) + 2 (x + 3)`
= `3(8 + 6) + 2(8 + 3)`
= `3(14) + 2(11)`
= `42 + 22`
= `64`
R.H.S.= `64`
L.H.S. = R.H.S.
Hence, verified.
20. `3(2 – 5x) – 2 (1 – 6x) = 1`
Solution:
`3(2 − 5x) − 2(1 − 6x) = 1`
⇒ `6 − 15x − 2 + 12x = 1`
⇒ `4 - 3x = 1`
⇒ `3 = 3x`
⇒ `x` = `frac\{cancel 3}{cancel 3}`
So, `x` = `1`
Verification:
Substituting `x = 1` in the L.H.S.:
L.H.S. = `3(2 – 5x) – 2 (1 – 6x) `
= `3(2 − 5 × 1) − 2(1 − 6 × 1)`
= `3(2 − 5) − 2(1− 6)`
= `3(−3) −2(−5)`
= `− 9 + 10`
= `1`
R.H.S.= `1`
So, L.H.S. = R.H.S.
Hence, verified.
21. `frac\{n}{4}`− 5 = `frac\{n}{6}` + `frac\{1}{2}`
Solution:
`frac\{n}{4}` − 5 = `frac\{n}{6}` + `frac\{1}{2}`
⇒ `frac\{n}{4}` − `frac\{n}{6}` = `frac\{1}{2}` + 5
[Transposing `frac\{n}{6}` to the L.H.S. and 5 to the R.H.S.]
⇒ `frac\{3n−2n}{12}` = `frac\{1+10}{2}`
⇒ `frac\{n}{12}` = `frac\{11}{2}`
⇒ `frac\{n}{cancel 12}` × `cancel 12` = `frac\{11}{cancel 2}` × `cancel 12^6`
[Dividing both the sides by 12]
⇒ `n = 66`
Verification:
Substituting `n = 66` on both the sides:
L.H.S. = `frac\{n}{4}`− `5`
= `frac\{cancel 66^33}{cancel 4^2}` − `5`
= `frac\{33}{2}` − `5`
= `frac\{33 − 10}{2}`
= `frac\{23}{2}`
R.H.S. = `frac\{n}{6}` + `frac\{1}{2}`
= `frac\{cancel 66^11}{cancel 6}` + `frac\{1}{2}`
= `11` + `frac\{1}{2}`
= `frac\{22 + 1}{2}`
= `frac\{23}{2}`
So, L.H.S. = R.H.S.
Hence, verified.
22. `frac\{2m}{3}` + 8 = `frac\{m}{2}` − 1
Solution:
`frac\{2m}{3}` + 8 = `frac\{m}{2}` − 1
⇒ `frac\{2m}{3}` − `frac\{m}{2}` = − 1 − 8
[Transposing `frac\{m}{2}` to the L.H.S. and 8 to the R.H.S.]
⇒ `frac\{4m−3m}{6}` = `− 9`
⇒ `frac\{m}{6}` = `− 9`
⇒ `frac\{m}{cancel 6}` × `cancel 6` = `− 9 × 6`
[Multiplying both the sides by 6]
⇒ `m = − 54`
Verification:
Substituting `x = −54` on both the sides:
L.H.S. = `frac\{2m}{3}` + `8`
= `frac\{2(−54)}{3}` + `8`
= `frac\{ − cancel 108^-36}{cancel 3}` + `8`
= `− 36 + 8`
= `− 28`
R.H.S.= `frac\{m}{2}` − `1`
= `frac\{− cancel 54^-27}{cancel 2}` − `1`
= `−27 − 1`
= `−28`
So, L.H.S. = R.H.S.
Hence, verified.
23. `frac\{2x}{5}` – `frac\{3}{2}` = `frac\{x}{2}` + `1`
Solution:
`frac\{2x}{5}` − `frac\{3}{2}` = `frac\{x}{2}` + `1`
⇒ `frac\{2x}{5}`− `frac\{x}{2}` = `1` + `frac\{3}{2}`
[Transposing `frac\{x}{2}` to the L.H.S. and `frac\{3}{2}` to R.H.S.]
⇒ `frac\{4x − 5x}{10}` = `frac\{2 + 3}{2}`
⇒ `frac\{− x}{10}` = `frac\{5}{2}`
⇒ `frac\{− x}{cancel 10}` × `cancel 10` = `frac\{5}{cancel 2}` × `cancel 10^5`
[Multiplying both the sides by 10]
⇒ `x = −25`
Verification:
Substituting `x = −25` on both the sides:
L.H.S. = `frac\{2x}{5}` – `frac\{3}{2}`
= `frac\{2(−25)}{5}` − `frac\{3}{2}`
= `frac\{−cancel 50^-10}{cancel 5}` − `frac\{3}{2}`
= `−10` − `frac\{3}{2}`
= `frac\{−23}{2}`
R.H.S. = `frac\{x}{2}` + `1`
= `frac\{−25}{2}`+ `1`
= `frac{−25 + 2}{2}`
= `frac\{−23}{2}`
So, L.H.S. = R.H.S.
Hence, verified.
24. `frac\{x−3}{5}` – 2 = `frac\{2x}{5}`
Solution:
`frac\{x−3}{5}` – 2 = `frac\{2x}{5}`
⇒ `frac\{x−3}{5}` – `cancel 2` + `cancel 2` = `frac\{2x}{5}` + `2`
[ Adding both side 2]
⇒ `frac\{x−3}{5}` – `frac\{2x}{5}` = `2`
⇒ `frac\{x −3 – 2x}{5}` = `2`
⇒ `frac\{ – x − 3}{5}` = `2`
⇒ `frac\{ – x − 3}{cancel 5}` × `cancel 5` = `2 × 5`
[ Multiply both side by 5]
⇒ ` – x − 3` = `10`
⇒ ` – x` = `10 + 3`
⇒ `x` = `− 13`
Verification:
Substituting `x = −13` on both the sides:
L.H.S. = `frac\{x−3}{5}` – `2`
= `frac\{−13 − 3}{5}` − `2`
= `frac\{−16}{5}` − `2`
= `frac\{−16 − 10}{5}`
= `frac\{−26}{5}`
R.H.S. = `frac\{2x}{5}`
= `frac\{2×(−13)}{5}`
= `frac\{−26}{5}`
L.H.S. = R.H.S.
Hence, verified.
25. `frac\{3x}{10}` – 4 = 14
Solution:
`frac\{3x}{10}` − `4` = `14`
⇒ `frac\{3x}{10}` − `cancel 4` + `cancel 4` = `14 + 4`
[Adding 4 on both the sides]
⇒ `frac\{3x}{10}` = 18
⇒ `frac\{3x}{cancel 10}` × `cancel 10` = `18 × 10`
[Multiplying both the sides by 10]
⇒ `3x = 180`
⇒ `frac\{cancel {3}x}{cancel 3}` = `frac\{cancel 180^60}{cancel 3}`
[Dividing both the sides by 3]
⇒ `x = 60`
Verification:
Substituting `x = 60` on both the sides:
L.H.S. = `frac\{3×60}{10}` − `4`
= `frac\{180}{10}` − `4`
= 18 − 4
= 14
R.H.S. = 14
So, L.H.S. = R.H.S.
Hence, verified.
26. `frac\{3}{4}``(x−1)` = `(x – 3)`
Solution:
`frac\{3}{4}``(x−1)` = `(x – 3)`
⇒`frac\{3x}{4}` − `frac\{3}{4}` = `x − 3`
⇒`frac\{3x}{4}` − `x` = ` − 3` + `frac\{3}{4}`
[Transposing x to the L.H.S. and `frac\{−3}{4}` to the R.H.S.]
⇒`frac\{3x−4x}{4}` = `frac\{−12 + 3}{4}`
⇒ `frac\{−x}{4}` = `frac {−9}{4}`
⇒ `frac\{−x}{cancel 4}` × `(−cancel 4)` = `frac\{−9}{cancel 4}` × `(−cancel 4)`
[Multiplying both the sides by -4]
⇒ `x = 9`
Verification:
Substituting `x = 9` on both the sides:
L.H.S. = `frac\{3}{4}``(x−1)`
= `frac\{3}{4}``(9 − 1)`
= `frac\{3}{cancel 4}` × `cancel 8^2`
= 6
R.H.S. = x − 3
= 9 − 3
= 6
So, L.H.S. = R.H.S.
Hence, verified.
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