Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RS Aggarwal Solutions: Linear Equations in One Variable
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Page 1
Q u e s t i o n : 1
3x - 5 = 0
S o l u t i o n :
3x -5 = 0 ? 3x = 5 Transposing -5 to RHS ? x =
5
3
CHECK: By substituting x =
5
3
in the given equation, we get: LHS = 3
5
3
- 5 = 5 - 5 = 0RHS = 0 ? LHS = RHS Hence
Q u e s t i o n : 2
8x - 3 = 9 - 2x
S o l u t i o n :
8x - 3 = 9 - 2x
?
8x + 2x = 9 + 3 Bytransposition
?
10x = 12
? x =
12
10
=
6
5
CHECK: By substituting x =
6
5
in the given equation, we get: LHS: 8
6
5
-3 =
48
5
-3 =
48-15
5
=
33
5
RHS: 9 -2
6
5
= 9 -
12
5
=
45-12
5
=
33
5
? LHS = RHS Hence checked.
Q u e s t i o n : 3
7 - 5x = 5 - 7x
S o l u t i o n :
We have:
7 - 5x = 5 - 7x
? -5x + 7x = 5 - 7 [transposing -7x to LHS and 7 to RHS]
? 2x = -2
? x =
-2
-1
2
1
? x = -1
Thus, x = -1 is a solution to the given equation.
CHECK: Substituting x = -1 in the given equation, we get:
LHS: = 7 - ?5x
= 7 -5 × (-1)
= 7 +5
= 12
RHS:
= 5 - 7x
=5 - 7 × (-1)
= 5 + 7
=12
? LHS = RHS
Hence, x = -1 is a solution of the given equation.
Q u e s t i o n : 4
3 + 2x = 1 - x
S o l u t i o n :
We have:
3+2x = 1 – x
? 2x + x+ 3 – 1 = 0 (By transposition)
? 3x + 2 = 0
? x = –
2
3
CHECK: Substituting x=-
2
3
in the given equation, we get:
LHS: 3+2x
=3+2 × -
2
3
=3 -
4
3
=
9-4
3
=
5
3
RHS: 1- x
=1-
-2
3
=1+
2
3
=
3+2
3
=
5
3
? LHS = RHS
Hence, x=-
2
3
is a solution of the given equation.
Q u e s t i o n : 5
2(x - 2) +3(4x - 1) = 0
S o l u t i o n :
( ) ( )
( ) ( )
( )
( )
Page 2
Q u e s t i o n : 1
3x - 5 = 0
S o l u t i o n :
3x -5 = 0 ? 3x = 5 Transposing -5 to RHS ? x =
5
3
CHECK: By substituting x =
5
3
in the given equation, we get: LHS = 3
5
3
- 5 = 5 - 5 = 0RHS = 0 ? LHS = RHS Hence
Q u e s t i o n : 2
8x - 3 = 9 - 2x
S o l u t i o n :
8x - 3 = 9 - 2x
?
8x + 2x = 9 + 3 Bytransposition
?
10x = 12
? x =
12
10
=
6
5
CHECK: By substituting x =
6
5
in the given equation, we get: LHS: 8
6
5
-3 =
48
5
-3 =
48-15
5
=
33
5
RHS: 9 -2
6
5
= 9 -
12
5
=
45-12
5
=
33
5
? LHS = RHS Hence checked.
Q u e s t i o n : 3
7 - 5x = 5 - 7x
S o l u t i o n :
We have:
7 - 5x = 5 - 7x
? -5x + 7x = 5 - 7 [transposing -7x to LHS and 7 to RHS]
? 2x = -2
? x =
-2
-1
2
1
? x = -1
Thus, x = -1 is a solution to the given equation.
CHECK: Substituting x = -1 in the given equation, we get:
LHS: = 7 - ?5x
= 7 -5 × (-1)
= 7 +5
= 12
RHS:
= 5 - 7x
=5 - 7 × (-1)
= 5 + 7
=12
? LHS = RHS
Hence, x = -1 is a solution of the given equation.
Q u e s t i o n : 4
3 + 2x = 1 - x
S o l u t i o n :
We have:
3+2x = 1 – x
? 2x + x+ 3 – 1 = 0 (By transposition)
? 3x + 2 = 0
? x = –
2
3
CHECK: Substituting x=-
2
3
in the given equation, we get:
LHS: 3+2x
=3+2 × -
2
3
=3 -
4
3
=
9-4
3
=
5
3
RHS: 1- x
=1-
-2
3
=1+
2
3
=
3+2
3
=
5
3
? LHS = RHS
Hence, x=-
2
3
is a solution of the given equation.
Q u e s t i o n : 5
2(x - 2) +3(4x - 1) = 0
S o l u t i o n :
( ) ( )
( ) ( )
( )
( )
We have:
2(x -2)+3(4x -1) = 0
? 2x -4 +12x -3 =0
? 14x -7 =0
? 14x = 7 (By transposition)
? x =
1
2
CHECK: Substituting x=
1
2
in the given equation, we get:
LHS: 2(x -2)+3(4x -1)
=2x -4 +12x -3
=2 ×
1
2
-4 +12 ×
1
2
-3
=1-4+6-3
=-7 +7
=0
RHS: 0
? LHS= RHS
Hence, x=
1
2
is a solution of the given equation.
Q u e s t i o n : 6
5(2x - 3) - 3(3x - 7) = 5
S o l u t i o n :
We have:
5(2x -3)-3(3x -7) = 5
? 10x -15 -9x +21 = 5
? 10x -9x =5+15 -21 (By transposition)
? x = 20 -21
? x = -1
CHECK: Substituting x=-1 in the given equation, we get:
LHS: 5(2x -3)-3(3x -7)
=10x -15 -9x +21
=10 ×(-1)-15 -9 ×(-1)+21
=-10 -15 +9 +21
=-25 +30
=5
RHS: 5
? LHS = RHS
Hence, x=-1 is a solution of the given equation.
Q u e s t i o n : 7
2x -
1
3
=
1
5
-x
S o l u t i o n :
We have:
2x -
1
3
=
1
5
-x
? 2x +x =
1
5
+
1
3
? 3x =
3×1+5×1
15
? 3x =
3+5
15
? 3x =
8
15
? x =
8
15×3
? x =
8
45
CHECK: Substituting x=
8
45
in the given equation, we get:
LHS: 2x -
1
3
=2 ×
8
45
-
1
3
=
16
45
-
1
3
=
16×1-15×1
45
=
16-15
45
=
1
45
RHS:
1
5
-x
=
1
5
-
8
45
=
1×9-1×8
45
=
9-8
45
=
1
45
? LHS=RHS
Hence, x=
8
45
is a solution of the given equation.
Q u e s t i o n : 8
1
2
x -3 = 5 +
1
3
x
S o l u t i o n :
Page 3
Q u e s t i o n : 1
3x - 5 = 0
S o l u t i o n :
3x -5 = 0 ? 3x = 5 Transposing -5 to RHS ? x =
5
3
CHECK: By substituting x =
5
3
in the given equation, we get: LHS = 3
5
3
- 5 = 5 - 5 = 0RHS = 0 ? LHS = RHS Hence
Q u e s t i o n : 2
8x - 3 = 9 - 2x
S o l u t i o n :
8x - 3 = 9 - 2x
?
8x + 2x = 9 + 3 Bytransposition
?
10x = 12
? x =
12
10
=
6
5
CHECK: By substituting x =
6
5
in the given equation, we get: LHS: 8
6
5
-3 =
48
5
-3 =
48-15
5
=
33
5
RHS: 9 -2
6
5
= 9 -
12
5
=
45-12
5
=
33
5
? LHS = RHS Hence checked.
Q u e s t i o n : 3
7 - 5x = 5 - 7x
S o l u t i o n :
We have:
7 - 5x = 5 - 7x
? -5x + 7x = 5 - 7 [transposing -7x to LHS and 7 to RHS]
? 2x = -2
? x =
-2
-1
2
1
? x = -1
Thus, x = -1 is a solution to the given equation.
CHECK: Substituting x = -1 in the given equation, we get:
LHS: = 7 - ?5x
= 7 -5 × (-1)
= 7 +5
= 12
RHS:
= 5 - 7x
=5 - 7 × (-1)
= 5 + 7
=12
? LHS = RHS
Hence, x = -1 is a solution of the given equation.
Q u e s t i o n : 4
3 + 2x = 1 - x
S o l u t i o n :
We have:
3+2x = 1 – x
? 2x + x+ 3 – 1 = 0 (By transposition)
? 3x + 2 = 0
? x = –
2
3
CHECK: Substituting x=-
2
3
in the given equation, we get:
LHS: 3+2x
=3+2 × -
2
3
=3 -
4
3
=
9-4
3
=
5
3
RHS: 1- x
=1-
-2
3
=1+
2
3
=
3+2
3
=
5
3
? LHS = RHS
Hence, x=-
2
3
is a solution of the given equation.
Q u e s t i o n : 5
2(x - 2) +3(4x - 1) = 0
S o l u t i o n :
( ) ( )
( ) ( )
( )
( )
We have:
2(x -2)+3(4x -1) = 0
? 2x -4 +12x -3 =0
? 14x -7 =0
? 14x = 7 (By transposition)
? x =
1
2
CHECK: Substituting x=
1
2
in the given equation, we get:
LHS: 2(x -2)+3(4x -1)
=2x -4 +12x -3
=2 ×
1
2
-4 +12 ×
1
2
-3
=1-4+6-3
=-7 +7
=0
RHS: 0
? LHS= RHS
Hence, x=
1
2
is a solution of the given equation.
Q u e s t i o n : 6
5(2x - 3) - 3(3x - 7) = 5
S o l u t i o n :
We have:
5(2x -3)-3(3x -7) = 5
? 10x -15 -9x +21 = 5
? 10x -9x =5+15 -21 (By transposition)
? x = 20 -21
? x = -1
CHECK: Substituting x=-1 in the given equation, we get:
LHS: 5(2x -3)-3(3x -7)
=10x -15 -9x +21
=10 ×(-1)-15 -9 ×(-1)+21
=-10 -15 +9 +21
=-25 +30
=5
RHS: 5
? LHS = RHS
Hence, x=-1 is a solution of the given equation.
Q u e s t i o n : 7
2x -
1
3
=
1
5
-x
S o l u t i o n :
We have:
2x -
1
3
=
1
5
-x
? 2x +x =
1
5
+
1
3
? 3x =
3×1+5×1
15
? 3x =
3+5
15
? 3x =
8
15
? x =
8
15×3
? x =
8
45
CHECK: Substituting x=
8
45
in the given equation, we get:
LHS: 2x -
1
3
=2 ×
8
45
-
1
3
=
16
45
-
1
3
=
16×1-15×1
45
=
16-15
45
=
1
45
RHS:
1
5
-x
=
1
5
-
8
45
=
1×9-1×8
45
=
9-8
45
=
1
45
? LHS=RHS
Hence, x=
8
45
is a solution of the given equation.
Q u e s t i o n : 8
1
2
x -3 = 5 +
1
3
x
S o l u t i o n :
We have:
1
2
x -3 = 5 +
1
3
x
?
1
2
x -
1
3
x = 5 +3 transposing
1
3
x to LHS and -3 to RHS
?
1×3-1×2
6
x = 8
?
3-2
6
x = 8
?
1
6
x = 8
? x = 8 × 6
? x = 48
CHECK: Substituting x=48 in the given equation, we get:
LHS:
1
2
x -3
=
1
2
1
×48
24
-3
=24-3
=21
RHS: 5 +
1
3
x
=5+
1
3
1
×48
16
=5+16
=21
? LHS=RHS
Hence, x=48 is a solution of the given equation.
Q u e s t i o n : 9
x
2
+
x
4
=
1
8
S o l u t i o n :
x
2
+
x
4
=
1
8
?
x×2+x×1
4
=
1
8
?
2x+x
4
=
1
8
?
3x
4
=
1
8
? 3x =
1
8
2
×4
1
? 3x =
1
2
? x =
1
6
CHECK: Substituting x=
1
6
in the given equation, we get:
LHS:
x
2
+
x
4
=
x×2+x×1
4
=
2x+x
4
=
3x
4
=
3
1
4
×
1
6
2
=
1
8
RHS:
1
8
? LHS = RHS
Hence, x=
1
3
is a solution of the given equation.
Q u e s t i o n : 1 0
3x + 2(x + 2) = 20 - (2x - 5)
S o l u t i o n :
( )
( )
( )
Page 4
Q u e s t i o n : 1
3x - 5 = 0
S o l u t i o n :
3x -5 = 0 ? 3x = 5 Transposing -5 to RHS ? x =
5
3
CHECK: By substituting x =
5
3
in the given equation, we get: LHS = 3
5
3
- 5 = 5 - 5 = 0RHS = 0 ? LHS = RHS Hence
Q u e s t i o n : 2
8x - 3 = 9 - 2x
S o l u t i o n :
8x - 3 = 9 - 2x
?
8x + 2x = 9 + 3 Bytransposition
?
10x = 12
? x =
12
10
=
6
5
CHECK: By substituting x =
6
5
in the given equation, we get: LHS: 8
6
5
-3 =
48
5
-3 =
48-15
5
=
33
5
RHS: 9 -2
6
5
= 9 -
12
5
=
45-12
5
=
33
5
? LHS = RHS Hence checked.
Q u e s t i o n : 3
7 - 5x = 5 - 7x
S o l u t i o n :
We have:
7 - 5x = 5 - 7x
? -5x + 7x = 5 - 7 [transposing -7x to LHS and 7 to RHS]
? 2x = -2
? x =
-2
-1
2
1
? x = -1
Thus, x = -1 is a solution to the given equation.
CHECK: Substituting x = -1 in the given equation, we get:
LHS: = 7 - ?5x
= 7 -5 × (-1)
= 7 +5
= 12
RHS:
= 5 - 7x
=5 - 7 × (-1)
= 5 + 7
=12
? LHS = RHS
Hence, x = -1 is a solution of the given equation.
Q u e s t i o n : 4
3 + 2x = 1 - x
S o l u t i o n :
We have:
3+2x = 1 – x
? 2x + x+ 3 – 1 = 0 (By transposition)
? 3x + 2 = 0
? x = –
2
3
CHECK: Substituting x=-
2
3
in the given equation, we get:
LHS: 3+2x
=3+2 × -
2
3
=3 -
4
3
=
9-4
3
=
5
3
RHS: 1- x
=1-
-2
3
=1+
2
3
=
3+2
3
=
5
3
? LHS = RHS
Hence, x=-
2
3
is a solution of the given equation.
Q u e s t i o n : 5
2(x - 2) +3(4x - 1) = 0
S o l u t i o n :
( ) ( )
( ) ( )
( )
( )
We have:
2(x -2)+3(4x -1) = 0
? 2x -4 +12x -3 =0
? 14x -7 =0
? 14x = 7 (By transposition)
? x =
1
2
CHECK: Substituting x=
1
2
in the given equation, we get:
LHS: 2(x -2)+3(4x -1)
=2x -4 +12x -3
=2 ×
1
2
-4 +12 ×
1
2
-3
=1-4+6-3
=-7 +7
=0
RHS: 0
? LHS= RHS
Hence, x=
1
2
is a solution of the given equation.
Q u e s t i o n : 6
5(2x - 3) - 3(3x - 7) = 5
S o l u t i o n :
We have:
5(2x -3)-3(3x -7) = 5
? 10x -15 -9x +21 = 5
? 10x -9x =5+15 -21 (By transposition)
? x = 20 -21
? x = -1
CHECK: Substituting x=-1 in the given equation, we get:
LHS: 5(2x -3)-3(3x -7)
=10x -15 -9x +21
=10 ×(-1)-15 -9 ×(-1)+21
=-10 -15 +9 +21
=-25 +30
=5
RHS: 5
? LHS = RHS
Hence, x=-1 is a solution of the given equation.
Q u e s t i o n : 7
2x -
1
3
=
1
5
-x
S o l u t i o n :
We have:
2x -
1
3
=
1
5
-x
? 2x +x =
1
5
+
1
3
? 3x =
3×1+5×1
15
? 3x =
3+5
15
? 3x =
8
15
? x =
8
15×3
? x =
8
45
CHECK: Substituting x=
8
45
in the given equation, we get:
LHS: 2x -
1
3
=2 ×
8
45
-
1
3
=
16
45
-
1
3
=
16×1-15×1
45
=
16-15
45
=
1
45
RHS:
1
5
-x
=
1
5
-
8
45
=
1×9-1×8
45
=
9-8
45
=
1
45
? LHS=RHS
Hence, x=
8
45
is a solution of the given equation.
Q u e s t i o n : 8
1
2
x -3 = 5 +
1
3
x
S o l u t i o n :
We have:
1
2
x -3 = 5 +
1
3
x
?
1
2
x -
1
3
x = 5 +3 transposing
1
3
x to LHS and -3 to RHS
?
1×3-1×2
6
x = 8
?
3-2
6
x = 8
?
1
6
x = 8
? x = 8 × 6
? x = 48
CHECK: Substituting x=48 in the given equation, we get:
LHS:
1
2
x -3
=
1
2
1
×48
24
-3
=24-3
=21
RHS: 5 +
1
3
x
=5+
1
3
1
×48
16
=5+16
=21
? LHS=RHS
Hence, x=48 is a solution of the given equation.
Q u e s t i o n : 9
x
2
+
x
4
=
1
8
S o l u t i o n :
x
2
+
x
4
=
1
8
?
x×2+x×1
4
=
1
8
?
2x+x
4
=
1
8
?
3x
4
=
1
8
? 3x =
1
8
2
×4
1
? 3x =
1
2
? x =
1
6
CHECK: Substituting x=
1
6
in the given equation, we get:
LHS:
x
2
+
x
4
=
x×2+x×1
4
=
2x+x
4
=
3x
4
=
3
1
4
×
1
6
2
=
1
8
RHS:
1
8
? LHS = RHS
Hence, x=
1
3
is a solution of the given equation.
Q u e s t i o n : 1 0
3x + 2(x + 2) = 20 - (2x - 5)
S o l u t i o n :
( )
( )
( )
We have:
3x +2(x +2) = 20 -(2x -5)
? 3x +2x +4 = 20 -2x +5
? 3x +2x +2x = 20 +5 -4 (Transposing -2x to LHS and 4 to RHS)
? 7x =21
? x =
21
3
7
1
? x = 3
CHECK: Substituting x=3 in the given equation, we get:
LHS=3x +2(x +2)
=3x +2x +4
=5x +4
=5 ×3+4
=15+4
=19
RHS=20 -(2x -5)
=20-2x +5
=25-2 ×3
=25-6
=19
? LHS = RHS
Hence, x=3 is a solution of the given equation.
Q u e s t i o n : 1 1
13(y - 4) - 3(y - 9) - 5(y + 4) = 0
S o l u t i o n :
We have:
13(y -4)-3(y -9)-5(y +4) = 0
? 13y -52 -3y +27 -5y -20 = 0
? 13y -3y -5y = 52 +20 -27 (Transposing -52, -20 and 27 to RHS)
? 5y =45
? y =
45
9
5
1
? y = 9
CHECK: Substituting x=9 in the given equation, we get:
LHS=13(y -4)-3(y -9)-5(y +4)
=13y-52 -3y +27 -5y -20
=13y-3y -5y -52 +27 -20
=5y-45
=5 ×9 -45
=45-45
=0
RHS=0
? LHS=RHS
Hence, x=9 is a solution of the given equation.
Q u e s t i o n : 1 2
2m+5
3
= 3m -10
S o l u t i o n :
We have,
2m+5
3
= 3m -10
? 2m +5 = 3(3m -10)
? 2m +5 = 9m -30
? 2m -9m = -30 -5 (Transposing 9m to LHS and 5 to RHS)
? -7m = -35
? m =
-35
5
-7
1
? m = 5
CHECK: Substituting m = 5 in the given equation, we get:
LHS=
2m+5
3
=
2×5+5
3
=
10+5
3
=
15
5
3
1
=5
RHS=3m -10
=3 ×5 -10
=15-10
=5
? LHS=RHS
Hence, x=5 is a solution of the given equation.
Q u e s t i o n : 1 3
6(3 x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x
S o l u t i o n :
Page 5
Q u e s t i o n : 1
3x - 5 = 0
S o l u t i o n :
3x -5 = 0 ? 3x = 5 Transposing -5 to RHS ? x =
5
3
CHECK: By substituting x =
5
3
in the given equation, we get: LHS = 3
5
3
- 5 = 5 - 5 = 0RHS = 0 ? LHS = RHS Hence
Q u e s t i o n : 2
8x - 3 = 9 - 2x
S o l u t i o n :
8x - 3 = 9 - 2x
?
8x + 2x = 9 + 3 Bytransposition
?
10x = 12
? x =
12
10
=
6
5
CHECK: By substituting x =
6
5
in the given equation, we get: LHS: 8
6
5
-3 =
48
5
-3 =
48-15
5
=
33
5
RHS: 9 -2
6
5
= 9 -
12
5
=
45-12
5
=
33
5
? LHS = RHS Hence checked.
Q u e s t i o n : 3
7 - 5x = 5 - 7x
S o l u t i o n :
We have:
7 - 5x = 5 - 7x
? -5x + 7x = 5 - 7 [transposing -7x to LHS and 7 to RHS]
? 2x = -2
? x =
-2
-1
2
1
? x = -1
Thus, x = -1 is a solution to the given equation.
CHECK: Substituting x = -1 in the given equation, we get:
LHS: = 7 - ?5x
= 7 -5 × (-1)
= 7 +5
= 12
RHS:
= 5 - 7x
=5 - 7 × (-1)
= 5 + 7
=12
? LHS = RHS
Hence, x = -1 is a solution of the given equation.
Q u e s t i o n : 4
3 + 2x = 1 - x
S o l u t i o n :
We have:
3+2x = 1 – x
? 2x + x+ 3 – 1 = 0 (By transposition)
? 3x + 2 = 0
? x = –
2
3
CHECK: Substituting x=-
2
3
in the given equation, we get:
LHS: 3+2x
=3+2 × -
2
3
=3 -
4
3
=
9-4
3
=
5
3
RHS: 1- x
=1-
-2
3
=1+
2
3
=
3+2
3
=
5
3
? LHS = RHS
Hence, x=-
2
3
is a solution of the given equation.
Q u e s t i o n : 5
2(x - 2) +3(4x - 1) = 0
S o l u t i o n :
( ) ( )
( ) ( )
( )
( )
We have:
2(x -2)+3(4x -1) = 0
? 2x -4 +12x -3 =0
? 14x -7 =0
? 14x = 7 (By transposition)
? x =
1
2
CHECK: Substituting x=
1
2
in the given equation, we get:
LHS: 2(x -2)+3(4x -1)
=2x -4 +12x -3
=2 ×
1
2
-4 +12 ×
1
2
-3
=1-4+6-3
=-7 +7
=0
RHS: 0
? LHS= RHS
Hence, x=
1
2
is a solution of the given equation.
Q u e s t i o n : 6
5(2x - 3) - 3(3x - 7) = 5
S o l u t i o n :
We have:
5(2x -3)-3(3x -7) = 5
? 10x -15 -9x +21 = 5
? 10x -9x =5+15 -21 (By transposition)
? x = 20 -21
? x = -1
CHECK: Substituting x=-1 in the given equation, we get:
LHS: 5(2x -3)-3(3x -7)
=10x -15 -9x +21
=10 ×(-1)-15 -9 ×(-1)+21
=-10 -15 +9 +21
=-25 +30
=5
RHS: 5
? LHS = RHS
Hence, x=-1 is a solution of the given equation.
Q u e s t i o n : 7
2x -
1
3
=
1
5
-x
S o l u t i o n :
We have:
2x -
1
3
=
1
5
-x
? 2x +x =
1
5
+
1
3
? 3x =
3×1+5×1
15
? 3x =
3+5
15
? 3x =
8
15
? x =
8
15×3
? x =
8
45
CHECK: Substituting x=
8
45
in the given equation, we get:
LHS: 2x -
1
3
=2 ×
8
45
-
1
3
=
16
45
-
1
3
=
16×1-15×1
45
=
16-15
45
=
1
45
RHS:
1
5
-x
=
1
5
-
8
45
=
1×9-1×8
45
=
9-8
45
=
1
45
? LHS=RHS
Hence, x=
8
45
is a solution of the given equation.
Q u e s t i o n : 8
1
2
x -3 = 5 +
1
3
x
S o l u t i o n :
We have:
1
2
x -3 = 5 +
1
3
x
?
1
2
x -
1
3
x = 5 +3 transposing
1
3
x to LHS and -3 to RHS
?
1×3-1×2
6
x = 8
?
3-2
6
x = 8
?
1
6
x = 8
? x = 8 × 6
? x = 48
CHECK: Substituting x=48 in the given equation, we get:
LHS:
1
2
x -3
=
1
2
1
×48
24
-3
=24-3
=21
RHS: 5 +
1
3
x
=5+
1
3
1
×48
16
=5+16
=21
? LHS=RHS
Hence, x=48 is a solution of the given equation.
Q u e s t i o n : 9
x
2
+
x
4
=
1
8
S o l u t i o n :
x
2
+
x
4
=
1
8
?
x×2+x×1
4
=
1
8
?
2x+x
4
=
1
8
?
3x
4
=
1
8
? 3x =
1
8
2
×4
1
? 3x =
1
2
? x =
1
6
CHECK: Substituting x=
1
6
in the given equation, we get:
LHS:
x
2
+
x
4
=
x×2+x×1
4
=
2x+x
4
=
3x
4
=
3
1
4
×
1
6
2
=
1
8
RHS:
1
8
? LHS = RHS
Hence, x=
1
3
is a solution of the given equation.
Q u e s t i o n : 1 0
3x + 2(x + 2) = 20 - (2x - 5)
S o l u t i o n :
( )
( )
( )
We have:
3x +2(x +2) = 20 -(2x -5)
? 3x +2x +4 = 20 -2x +5
? 3x +2x +2x = 20 +5 -4 (Transposing -2x to LHS and 4 to RHS)
? 7x =21
? x =
21
3
7
1
? x = 3
CHECK: Substituting x=3 in the given equation, we get:
LHS=3x +2(x +2)
=3x +2x +4
=5x +4
=5 ×3+4
=15+4
=19
RHS=20 -(2x -5)
=20-2x +5
=25-2 ×3
=25-6
=19
? LHS = RHS
Hence, x=3 is a solution of the given equation.
Q u e s t i o n : 1 1
13(y - 4) - 3(y - 9) - 5(y + 4) = 0
S o l u t i o n :
We have:
13(y -4)-3(y -9)-5(y +4) = 0
? 13y -52 -3y +27 -5y -20 = 0
? 13y -3y -5y = 52 +20 -27 (Transposing -52, -20 and 27 to RHS)
? 5y =45
? y =
45
9
5
1
? y = 9
CHECK: Substituting x=9 in the given equation, we get:
LHS=13(y -4)-3(y -9)-5(y +4)
=13y-52 -3y +27 -5y -20
=13y-3y -5y -52 +27 -20
=5y-45
=5 ×9 -45
=45-45
=0
RHS=0
? LHS=RHS
Hence, x=9 is a solution of the given equation.
Q u e s t i o n : 1 2
2m+5
3
= 3m -10
S o l u t i o n :
We have,
2m+5
3
= 3m -10
? 2m +5 = 3(3m -10)
? 2m +5 = 9m -30
? 2m -9m = -30 -5 (Transposing 9m to LHS and 5 to RHS)
? -7m = -35
? m =
-35
5
-7
1
? m = 5
CHECK: Substituting m = 5 in the given equation, we get:
LHS=
2m+5
3
=
2×5+5
3
=
10+5
3
=
15
5
3
1
=5
RHS=3m -10
=3 ×5 -10
=15-10
=5
? LHS=RHS
Hence, x=5 is a solution of the given equation.
Q u e s t i o n : 1 3
6(3 x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x
S o l u t i o n :
We have:
6(3x +2)-5(6x -1) = 3(x -8)-5(7x -6)+9x
? 18x+12 -30x +5 =3x -24 -35x +30 +9x
? 18x -30x -3x +35x -9x = -24 +30 -12 -5 (Transposing 3x, 9x and -35x to LHS and 12 and 5 to RHS)
? 53x -42x =30 -41
? 11x = -11
? x =
-11
1
11
1
? x = -1
CHECK: Substituting x=-1 in the given equation, we get:
LHS=6(3x +2)-5(6x -1)
=18x+12 -30x +5
=-12x +17
=-12 ×(-1)+17
=12+17
=29
RHS=3(x -8)-5(7x -6)+9x
=3x -24 -35x +30 +9x
=12x -35x -24 +30
=-23x +6
=-23 ×(-1)+6
= 23+6
=29
? LHS=RHS
Hence, x=-1 is a solution of the given equation.
Q u e s t i o n : 1 4
t - (2t + 5) - 5(1 - 2t) = 2(3 + 4t) -3(t - 4)
S o l u t i o n :
We have:
t -(2t +5)-5(1 -2t) = 2(3 +4t)-3(t -4)
? t -2t -5 -5 +10t =6+8t -3t +12
? t -2t +10t -8t +3t = 6 +12 +5 +5 (By transposition )
? 14t -10t =28
? 4t = 28
? x =
28
7
4
1
? x = 7
CHECK: Substituting x=7 in the given equation, we get:
LHS=t -(2t +5)-5(1 -2t)
=t -2t -5 -5 +10t
=11t -2t -10
=9t -10
=9 ×7 -10
=63-10
=53
RHS=2(3 +4t)-3(t -4)
=6+8t -3t +12
=5t +18
=5 ×7 +18
=35+18
= 53
? LHS=RHS
Hence, x=7 is a solution of the given equation.
Q u e s t i o n : 1 5
2
3
x =
3
8
x +
7
12
S o l u t i o n :
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FAQs on RS Aggarwal Solutions: Linear Equations in One Variable - Mathematics (Maths) Class 7
| 1. What are linear equations in one variable? |  |
Ans. Linear equations in one variable are equations that involve only one variable and have a degree of 1. They can be written in the form ax + b = 0, where 'a' and 'b' are constants and 'x' is the variable. The solution to a linear equation in one variable is a single value of the variable that satisfies the equation.
| 2. How do you solve linear equations in one variable? | |
Ans. To solve a linear equation in one variable, follow these steps:
1. Simplify both sides of the equation by performing any necessary operations.
2. Isolate the variable term on one side of the equation.
3. Perform inverse operations to get the variable term alone.
4. Check the solution by substituting it back into the original equation.
| 3. Can linear equations in one variable have multiple solutions? | |
Ans. No, linear equations in one variable can have at most one solution. This means that there is either one unique solution that satisfies the equation, or there are no solutions at all. If the equation has no solution, it is called inconsistent. If the equation has infinitely many solutions, it is called dependent.
| 4. Are all equations in the form ax + b = 0 considered linear equations in one variable? | |
Ans. Yes, all equations in the form ax + b = 0 are considered linear equations in one variable. This is because these equations involve only one variable, 'x', and have a degree of 1. The 'a' and 'b' in the equation are constants that determine the coefficients of the variable.
| 5. How are linear equations in one variable used in real-life situations? | |
Ans. Linear equations in one variable are used in various real-life situations, such as:
- Calculating the cost of a product based on the number of items purchased.
- Determining the time taken to travel a certain distance at a given speed.
- Estimating the growth rate of a population over time.
- Analyzing the relationship between two variables in a scientific experiment.
- Solving problems involving proportions and ratios.
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