Solve Equations by Substitutions

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

Examples with Solutions

Example 1

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

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

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

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

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.

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