Example 1
Solve the equation
x  3 √x =  2
Solution to Example 1

Let u = √x) so that u^{ 2} = x. We now substitute x and √ (x) by u and u^{ 2} respectively to obtain an equation in u.
u^{ 2}  3 u =  2

The above is a quadratic equation; rewrite it so that its right term is equal to zero.
u^{ 2}  3 u + 2 = 0

Use any method to solve the above equation for u to obtain:
u = 1 or u = 2

We now substitute u by √x and solve for x
√x = 1 or √x = 2
x = 1 or x = 4
Example 2
Use the method of substitution to solve the equation
1 / (x  1)^{ 2}  1 / (x  1)  2 = 0
Solution to Example 2

Let u = 1 / (x  1) and make the substitution in the above equation to obtain an equation in u.
u^{ 2}  u  2 = 0

Solve the above quadratic equation to obtain:
u =  1 or u = 2

We now substitute u by 1 / (x  1) and solve for x
1 / (x  1) =  1 or 1 / (x  1) = 2
x = 0 or x = 3 / 2
Example 3:
Use the method of substitution to solve the equation
 (x + 3)^{ 6} + 4 (x + 3)^{ 3} =  21
Solution to Example 3

Let u = (x + 3)^{ 3} and substitute in the above equation to obtain an equation in u.
 u^{ 2} + 4 u = 21
u^{ 2}  4 u  21 = 0

Solve the above quadratic equation to obtain:
u =  3 or u = 7

We now substitute u by (x + 3)^{ 3} and solve for x
(x + 3)^{ 3} =  3 or (x + 3)^{ 3} = 7

We now solve for x
x =  3  cube root ( 3 )
or
x =  3 + cube root ( 7 )
Example 4
Use the method of substitution to solve the equation
3 e^{ 2 x}  e^{ x}  2 = 0
Solution to Example 4

Let u = e^{ x} so that u^{ 2} = e^{ 2 x} and substitute in the above equation to obtain:
3 u^{ 2}  u  2 = 0

Use any method to solve the above quadratic equation.
u = 1
or
u =  2 / 3

We now substitute u by e^{ x} and solve for x
e^{ x} = 1 or e^{ x} =  2 / 3

We now solve e^{ x} = 1 to obtain:
x = 0

The second equation e^{ x} =  2 / 3 has no solution since e^{ x} is always positive.
Example 5
Use the method of substitution to solve the equation
sin^{ 2} x  4 sin x  5 = 0
for x in the interval [ 0 , 2 π ). Give x in radians.
Solution to Example 5

Let u = sin x and substitute in the above equation to obtain:
u^{ 2}  4 u  5 = 0

Solve the above quadratic equation.
u =  1
or
u = 5

We now substitute u by sin x and solve for x
sin x =  1 or sin x = 5

We solve sin x =  1 to obtain:
x = 3 π / 2

The range of sin x is the set of values in the interval [  1 , 1 ] and therefore the equation sin x = 5 has no solution.
More References and linksSolve Equations, Systems of Equations and Inequalities.
