RS Aggarwal Solutions for Class 7 Maths Exercise 7A Chapter 7 Linear Equations in One Variable, an equation is a statement of equality which contains one or more unknown quantities or variables. An equation involving only a linear polynomial is a linear equation. Students of Class 7 are suggested to solve RS Aggarwal Class 7 Solutions Chapter 7 to strengthen the fundamentals and be able to solve questions that are usually asked in the examination. By practising RS Aggarwal Solutions for Class 7 Maths Chapter 7 Linear Equations in One Variable, students can get good marks in Maths.
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Solve the following equations. Check your result in each case.
- 3x â€“ 5 = 0
Solution:-
The value of the variable which makes the equation a true statement is called the root of the equation.
Now, find the value of x by considering the given equation,
= 3x â€“ 5 = 0
Transposing â€“ 5 to RHS it becomes 5
= 3x = 5
Multiplying both side by (1/3)
= 3x Ã— (1/3) = 5 Ã— (1/3)
= x = (5/3)
By substituting (5/3) in the place of x in given equation, we get
= 3 Ã— (5/3) â€“ 5 = 0
= 5 â€“ 5 = 0
= 0= 0
âˆ´ LHS = RHS
Hence, the result is verified.
- 8x â€“ 3 = 9 â€“ 2x
Solution:-
Now, find the value of x by considering the given equation,
=8x â€“ 3 = 9 â€“ 2x
Transposing â€“ 3 to RHS and it becomes 3 and -2x to LHS it becomes 2x
= 8x + 2x = 9 + 3
= 10x = 11
Multiplying both side by (1/10)
= 10x Ã— (1/10) = 12 Ã— (1/10)
= x = (12/10)
= x = (6/5)
By substituting (6/5) in the place of x in given equation, we get
LHS,
= 8 Ã— (6/5) – 3
= (48/5) – 3
= (48-15/5)
= (33/5)
RHS,
= 9 â€“ 2 Ã— (6/5)
= 9 â€“ (12/5)
= (45 â€“ 12)/5
= (33/5)
By comparing LHS and RHS
= (33/5) = (33/5)
âˆ´ LHS = RHS
Hence, the result is verified.
- 7 â€“ 5x = 5 â€“ 7x
Solution:-
Now, find the value of x by considering the given equation,
=7 â€“ 5x = 5 â€“ 7x
Transposing 7 to RHS and it becomes -7 and -7x to LHS it becomes 7x
= 7x â€“ 5x = 5 – 7
= 2x = -2
Multiplying both side by (1/3)
= 2x Ã— (1/2) = -2 Ã— (1/2)
= x = (-2/2)
= x = -1
By substituting -1 in the place of x in given equation, we get
LHS,
= 7 â€“ 5 Ã— (-1)
= 7 â€“ (-5)
= 7 + 5
= 12
RHS,
= 5 â€“ 7 Ã— (-1)
= 5 â€“ (-7)
= 5 + 7
= 12
By comparing LHS and RHS
= 12 = 12
âˆ´ LHS = RHS
Hence, the result is verified.
- 3 + 2x = 1 â€“ x
Solution:-
Now, find the value of x by considering the given equation,
=3 + 2x = 1 â€“ x
Transposing 3 to RHS and it becomes -3 and -x to LHS it becomes x
= 2x + x = 1 – 3
= 3x = -2
Multiplying both side by (1/3)
= 3x Ã— (1/3) = -2 Ã— (1/3)
= x = (-2/3)
By substituting -2/3 in the place of x in given equation, we get
LHS,
= 3 + 2 Ã— (-2/3)
= 3 + (-4/3)
= 3 – (4/3)
= (9 – 4) / 3
= 5/3
RHS,
= 1 â€“ (-2/3)
= 1 + (2/3)
= (3 + 2)/ 3
= 5/3
By comparing LHS and RHS
= 5/3 = 5/3
âˆ´ LHS = RHS
Hence, the result is verified.
- 2 (x â€“ 2) + 3 (4x â€“ 1) = 0
Solution:-
The above question can be written as,
= 2x â€“ 4 + 12x â€“ 3 = 0
Now, find the value of x by considering the above equation,
= 2x + 12x â€“ 4 â€“ 3 = 0
= 14x â€“ 7 = 0
Transposing -7 to RHS and it becomes 7
= 14x = 7
Multiplying both side by (1/14)
= 14x Ã— (1/14) = 7 Ã— (1/14)
= x = (7/14)
= x = 1/2
By substituting 1/2 in the place of x in given equation, we get
LHS,
= (2 Ã— (1/2)) â€“ 4 + (12 Ã— (1/2)) â€“ 3
= 1â€“ 4 + 6 â€“ 3
= 7 â€“ 7
= 0
By comparing LHS and RHS
= 0 = 0
âˆ´ LHS = RHS
Hence, the result is verified.
- 5 (2x â€“ 3) – 3 (3x â€“ 7) = 5
Solution:-
The above question can be written as,
= 10x â€“ 15 – 9x + 21 = 5
Now, find the value of x by considering the given equation,
= 10x – 15 â€“ 9x + 21 = 5
= x + 6 = 5
Transposing 6 to RHS and it becomes -6
= x = 5 â€“ 6
= x = -1
By substituting -1 in the place of x in given equation, we get
LHS,
= 10 (-1) -15 -9(-1) + 21
= -10 â€“ 15 + 9 + 21
= -25 + 30
= 5
By comparing LHS and RHS
= 5 = 5
âˆ´ LHS = RHS
Hence, the result is verified.
- 2x â€“ (1/3) = (1/5) â€“ x
Solution:-
Now, find the value of x by considering the given equation,
=2x â€“ (1/3) = (1/5) â€“ x
Transposing â€“ (1/3) to RHS and it becomes (1/3) and – x to LHS it becomes x
= 2x + x = (1/5) + (1/3)
= 3x = (3 + 5) / 15
= 3x = 8/15
Multiplying both side by (1/3)
= 3x Ã— (1/3) = (8/15) Ã— (1/3)
= x = (8/45)
By substituting (8/45) in the place of x in given equation, we get
LHS,
= 2 Ã— (8/45) â€“ (1/3)
= (16/45) â€“ (1/3)
= (16 â€“ 15) / 45
= (1/45)
RHS
= (1/5) â€“ (8/45)
= (9-8) / 45
= (1/45)By comparing LHS and RHS
= (1/45) = (1/45)
âˆ´ LHS = RHS
Hence, the result is verified.
- (1/2)x- 3 = 5 + (1/3)x
Solution:-
Now, find the value of x by considering the given equation,
= (1/2)x- 3 = 5 + (1/3)x
Transposing -3 to RHS and it becomes 3 and (1/3)x to LHS it becomes (-1/3)x
= (1/2)x – (1/3)x = 5 + 3
= ((3-2)/6)x = 8
= (1/6)x = 8
Multiplying both side by 6
= 6 Ã— (1/6)x = 8 Ã— 6
= x = 48
By substituting 48 in the place of x in given equation, we get
LHS,
= (1/2) Ã— (48) – 3
= (48/2) – 3
= 24 – 3
= 21
RHS,
= 5 + (1/3) Ã— (48)
= 5 + (48/3)
= 5 + 16
= 21
By comparing LHS and RHS
= 21 = 21
âˆ´ LHS = RHS
Hence, the result is verified.
- (x/2) + (x/4) = (1/8)
Solution:-
Now, find the value of x by considering the given equation,
= (x/2) + (x/4) = (1/8)
= ((2 + 1) / 4) = (1/8)
= (3/4)x = (1/8)
Multiplying both side by (4/3)
= (3/4)x Ã— (4/3) = (1/8) Ã— (4/3)
= x = (1/6)
By substituting (1/6) in the place of x in given equation, we get
LHS,
= (3/4)x
= (3/4) Ã— (1/6)
= (1/8)
RHS,
= (1/8)
By comparing LHS and RHS
= (1/8) = (1/8)
âˆ´ LHS = RHS
Hence, the result is verified.
- 3x + 2(x + 2) = 20 â€“ (2x â€“ 5)
Solution:-
The above question can be written as,
= 3x + 2x + 4 = 20 â€“ 2x + 5
Now, find the value of x by considering the above equation,
= 5x + 4 = 20 â€“ 2x + 5
Transposing 4 to RHS and it becomes -4 and -2x to LHS it becomes 2x
= 5x + 2x = 20 â€“ 4 + 5
= 7x = 25 â€“ 4
= 7x = 21
Multiplying both side by (1/7)
= 7x Ã— (1/7) = 21 Ã— (1/7)
= x = 21/7
= x = 3
By substituting 3 in the place of x in given equation, we get
LHS,
= 5x + 4
= 5 Ã— (3) + 4
= 15 + 4
= 19
RHS,
= 20 â€“ 2x + 5
= 20 â€“ 2 Ã— (3) + 5
= 20 â€“ 6 + 5
= 25 â€“ 6
= 19
By comparing LHS and RHS
= 19 = 19
âˆ´ LHS = RHS
Hence, the result is verified.
- 13( y â€“ 4) â€“ 3(y-9) â€“ 5(y + 4) = 0
Solution:-
The above question can be written as,
= 13y â€“ 52 â€“ 3y + 27 â€“ 5y â€“ 20 = 0
Now, find the value of y by considering the above equation,
= 13y â€“ 3y â€“ 5y â€“ 52 + 27 â€“ 20 = 0
= 13y â€“ 8y â€“ 72 + 27 = 0
= 5y â€“ 45 = 0
Transposing -45 to RHS and it becomes 45
= 5y = 45
Multiplying both side by (1/5)
= 5y Ã— (1/5) = 45 Ã— (1/5)
= y = 9
By substituting 9 in the place of y in given equation, we get
LHS,
= 5y â€“ 45
= 5 Ã— (9) â€“ 45
= 45 â€“ 45
= 0
RHS,
= 0
By comparing LHS and RHS
= 0 = 0
âˆ´ LHS = RHS
Hence, the result is verified.
- (2m + 5) / 3 = (3m â€“ 10)
Solution:-
The above question can be written as,
= 3 Ã— [(2m + 5)/ 3] = 3 Ã— (3m -10) â€¦ [Multiplying both side by 3]
= (2m + 5) = 9m â€“ 30
Transposing 2m to RHS and it becomes -2m and -30 to LHS it becomes 30
= 30 + 5 = 9m -2m
= 35 = 7m
Multiplying both side by (1/7),
= 7m Ã— (1/7) = 35 Ã— (1/7)
= m = (35/7)
= m = 5
By substituting 5 in the place of m in given equation, we get
LHS,
= 2m + 5
= 2 Ã— (5) + 5
= 10 + 5
= 15
RHS,
= 9m â€“ 30
= 9 Ã— (5) â€“ 30
= 45 â€“ 30
= 15
By comparing LHS and RHS
= 15 = 15
âˆ´ LHS = RHS
Hence, the result is verified.
- 6(3x + 2) â€“ 5(6x â€“ 1) = 3(x â€“ 8) â€“ 5(7x â€“ 6) + 9x
Solution:-
The above question can be written as,
= 18x + 12 â€“ 30x + 5 = 3x â€“ 24 â€“ 35x + 30 + 9x
Now, find the value of x by considering the above equation,
Transposing 12, 5 to RHS and it becomes -12, -5 and -35x, 9x, 3x to LHS it becomes 35x, -9x, -3x
= 18x – 30x + 35x â€“ 9x – 3x = -24 + 30 â€“ 12 – 5
= 53x â€“ 42x = 30 â€“ 41
= 11x = -11
Multiplying both side by (1/11),
= 11x Ã— (1/11) = -11 Ã— (1/11)
= x = (-11/11)
= x = -1
By substituting 5 in the place of x in given equation, we get
LHS,
= 18x + 12 â€“ 30x + 5
= 18 (-1) + 12 â€“ 30 (-1) + 5
= -18 + 12 + 30 + 5
= -18 + 47
= 29
RHS,
= 3x â€“ 24 â€“ 35x + 30 + 9x
= 3 (-1) â€“ 24 â€“ 35 (-1) + 30 + 9 (-1)
= -3 â€“ 24 + 35 + 30 â€“ 9
= -36 + 65
= 29
By comparing LHS and RHS
= 15 = 15
âˆ´ LHS = RHS
Hence, the result is verified.
- t â€“ (2t + 5) â€“ 5(1 â€“ 2t) = 2(3 + 4t) â€“ 3(t -4)
Solution:-
The above question can be written as,
= t â€“ 2t â€“ 5 â€“ 5 + 10t = 6 + 8t â€“ 3t + 12
Now, find the value of t by considering the above equation,
Transposing -5, -5 to RHS and it becomes 5, 5 and 8t, -3t to LHS it becomes -8t, 3t
= t â€“ 2t + 10t â€“ 8t + 3t = 6 + 12 + 5 + 5
= 14t â€“ 10t = 28
= 4t = 28
Multiplying both side by (1/4),
= 4t Ã— (1/4) = 28 Ã— (1/4)
= t = 7
By substituting 7 in the place of t in given equation, we get
LHS,
= t â€“ 2t â€“ 5 â€“ 5 + 10t
= 7 – 2Ã— (7) â€“ 5 â€“ 5 + 10(7)
= 7 â€“ 14 â€“ 5 â€“ 5 + 70
= 77 â€“ 24
= 53
RHS,
= 6 + 8t â€“ 3t + 12
= 6 + 8 Ã— (7) â€“ 3 Ã— (7) + 12
= 6 + 56 â€“ 21 + 12
= 74 â€“ 21
= 53
By comparing LHS and RHS
= 15 = 15
âˆ´ LHS = RHS
Hence, the result is verified.
- (2/3)x = (3/8)x + (7/12)
Solution:-
Find the value of x by considering the above equation,
Transposing (3/8)x to RHS and it becomes -(3/8)x.
= (2/3)x â€“ (3/8)x = (7/12)
= [(16 â€“ 9) / 24]x = (7/12)
= (7/24) x = (7/12)
Multiplying both side by (24/7),
= (7/24)x Ã— (24/7) = (7/12) Ã— (24/7)
= x = 2
By substituting 2 in the place of x in given equation, we get
LHS,
= (2/3)x
= (2/3) Ã— (2)
= (4/3)
RHS,
= (3/8)x + (7/12)
= (3/8) Ã— (2) + (7/12)
= (3/4) + (7/12)
= (9 + 7) / (12)
= (16/12)
= (4/3)
By comparing LHS and RHS
= (4/3) = (4/3)
âˆ´ LHS = RHS
Hence, the result is verified.
- [(3x â€“ 1)/ 5] â€“ (x/7) = 3
Solution:-
The above question can be written as,
= [(7 Ã— (3x -1)) â€“ (5 Ã— (x))/35] = 3
= ((21x â€“ 7) â€“ (5x))/35 = 3
= (21x â€“ 7 â€“ 5x)/35 = 3
= (16x â€“ 7)/35 = 3 â€¦ [Multiplying both side by 35]
= [(16x â€“ 7)/35] Ã— 35 = 3 Ã— 35
= 16x â€“ 7 = 105
= 16x = 105 + 7
= 16x = 112 â€¦ [Multiplying both side by 1/16]
= 16x Ã— (1/16) = 112 Ã— (1/16)
= x = 7
By substituting 7 in the place of x in given equation, we get
LHS,
= (16x â€“ 7)/35
= ((16 Ã— 7) â€“ 7)/35
= (112 â€“ 7)/ 35
= (105/35)
= 3
RHS,
= 3
By comparing LHS and RHS
= (4/3) = (4/3)
âˆ´ LHS = RHS
Hence, the result is verified.