Linear equation in one variable:
An equation of one variable and first order is called a
linear equation in one variable such an equation has
only one solution. A solution is also called the ‘root’ of
the given equation. An equation that can be written in
the form,
ax + b = 0,
Where a and b are real numbers and a ≠ o.
Following are some examples:
a) 2x + 3 = 0
x 1 2
b)
3 2 3
+ =
c) 5(3x+2) - 2 = - 2(1-7x)
Example-1
Solve the equation x – 4 = 5.
Solution:
Since x – 4 = 5
x = 5 + 4
x = 9
Example-2
Solve the equation 3x – 4 = 5.
Solution:
Since 3x – 4 = 5
3x = 5 + 4
3x = 9
9
x
3
=
x = 3
Example-3
Solve the equation 3-{2(1-x)-x} = 4
Solution:
Since 3-{2(1-x)-x} = 4
3-{2-2x-x} = 4
3-(2-3x) = 4
3-2+3x = 4
1+3x = 4
3x = 4-1
3x = 3
3
x
3
=
x = 1
Example-4
Solve the equation
x 5 x 3 5
7 4 14
+ −
+ =
Solution:
x 5 x 3 5
since
7 4 14
4(x 5) 7(x 3) 5
28 14
4x 20 7x 21
5
2
+ −
+ =
+ + −
=
+ + −
=
11x 1
5
2
11x 1 5x 2
11x 1 10
11x 10 1
11x 11
11
x
11
x 1
−
=
− =
− =
= +
=
=
=
Example-5
Solve the equation
x (x 3)
1
4 6
−
− =
Solution:
x (x 3)
1
4 6
multiplying throghout by the LCM (2 x 2 x 3) of the denominators (which is 12)
x 3x
12 12 12(1)
64
3x 2(x 3) 12
3x 2x 6 12
x 6 12
x 12 6
x 6
−
− =
−  − =   
   
− − =
− + =
+ =
= −
=

Linear equation in one variable

  • 2.
    Linear equation inone variable: An equation of one variable and first order is called a linear equation in one variable such an equation has only one solution. A solution is also called the ‘root’ of the given equation. An equation that can be written in the form, ax + b = 0, Where a and b are real numbers and a ≠ o. Following are some examples: a) 2x + 3 = 0 x 1 2 b) 3 2 3 + = c) 5(3x+2) - 2 = - 2(1-7x)
  • 3.
    Example-1 Solve the equationx – 4 = 5. Solution: Since x – 4 = 5 x = 5 + 4 x = 9 Example-2 Solve the equation 3x – 4 = 5. Solution: Since 3x – 4 = 5 3x = 5 + 4 3x = 9 9 x 3 = x = 3
  • 4.
    Example-3 Solve the equation3-{2(1-x)-x} = 4 Solution: Since 3-{2(1-x)-x} = 4 3-{2-2x-x} = 4 3-(2-3x) = 4 3-2+3x = 4 1+3x = 4 3x = 4-1 3x = 3 3 x 3 = x = 1
  • 5.
    Example-4 Solve the equation x5 x 3 5 7 4 14 + − + = Solution: x 5 x 3 5 since 7 4 14 4(x 5) 7(x 3) 5 28 14 4x 20 7x 21 5 2 + − + = + + − = + + − = 11x 1 5 2 11x 1 5x 2 11x 1 10 11x 10 1 11x 11 11 x 11 x 1 − = − = − = = + = = =
  • 6.
    Example-5 Solve the equation x(x 3) 1 4 6 − − = Solution: x (x 3) 1 4 6 multiplying throghout by the LCM (2 x 2 x 3) of the denominators (which is 12) x 3x 12 12 12(1) 64 3x 2(x 3) 12 3x 2x 6 12 x 6 12 x 12 6 x 6 − − = −  − =        − − = − + = + = = − =