Solve Equations by Substitution

The powerful method of substitution is used to solve different types of equations.

Example 1: Solve the equation

x - 3 sqrt ( x ) = - 2

Solution to Example 1:

  • Let u = sqrt ( x ) so that u 2 = x. We now substitute x and sqrt (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 sqrt ( x ) and solve for x
    sqrt ( x ) = 1 or sqrt ( x ) = 2

    x = 1 or x = 4

Example 2: Use the method of sustitution 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 sustitution 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 sustitution 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 sustitution to solve the equation

sin 2 x - 4 sin x - 5 = 0
for x in the interval [ 0 , 2 Pi ). 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 Pi / 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 links on how to Solve Equations, Systems of Equations and Inequalities.

privacy policy