Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RS Aggarwal Solutions: Linear Equations in One Variable
RS Aggarwal Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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 :
Read More
|
76 videos|345 docs|39 tests
|
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.
Related Exams
About this Document
|
4.67/5 Rating |
|
Nov 21, 2024 Last updated |
Document Description: RS Aggarwal Solutions: Linear Equations in One Variable for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RS Aggarwal Solutions: Linear Equations in One Variable have been prepared according to the Class 7 exam syllabus. Information about RS Aggarwal Solutions: Linear Equations in One Variable covers topics
like and RS Aggarwal Solutions: Linear Equations in One Variable Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Linear Equations in One Variable.
Introduction of RS Aggarwal Solutions: Linear Equations in One Variable in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RS Aggarwal Solutions: Linear Equations in One Variable in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: RS Aggarwal Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Linear Equations in One Variable
In this doc you can find the meaning of RS Aggarwal Solutions: Linear Equations in One Variable defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal Solutions: Linear Equations in One Variable theory, EduRev gives you an ample number of questions to practice RS Aggarwal Solutions: Linear Equations in One Variable tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Extra Questions
,Summary
,RS Aggarwal Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 7
,MCQs
,RS Aggarwal Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 7
,mock tests for examination
,shortcuts and tricks
,Exam
,Sample Paper
,study material
,video lectures
,ppt
,past year papers
,Previous Year Questions with Solutions
,Important questions
,Viva Questions
,RS Aggarwal Solutions: Linear Equations in One Variable | Mathematics (Maths) Class 7
,Semester Notes
,practice quizzes
,Objective type Questions
,Free
;
Additional Information about RS Aggarwal Solutions: Linear Equations in One Variable for Class 7 Preparation
RS Aggarwal Solutions: Linear Equations in One Variable Free PDF Download
The RS Aggarwal Solutions: Linear Equations in One Variable is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RS Aggarwal Solutions: Linear Equations in One Variable now and kickstart your journey towards success in the Class 7 exam.
Importance of RS Aggarwal Solutions: Linear Equations in One Variable
The importance of RS Aggarwal Solutions: Linear Equations in One Variable cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RS Aggarwal Solutions: Linear Equations in One Variable Notes
RS Aggarwal Solutions: Linear Equations in One Variable Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Linear Equations in One Variable.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RS Aggarwal Solutions: Linear Equations in One Variable Notes on EduRev are your ultimate resource for success.
RS Aggarwal Solutions: Linear Equations in One Variable Class 7 Questions
The "RS Aggarwal Solutions: Linear Equations in One Variable Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RS Aggarwal Solutions: Linear Equations in One Variable on the App
Students of Class 7 can study RS Aggarwal Solutions: Linear Equations in One Variable alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Linear Equations in One Variable,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RS Aggarwal Solutions: Linear Equations in One Variable is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily