Hints for solving the problems(8.3)

Solving Equations with Rational Expressions

To solve any of the following equations you need to determine the LCD, and multiply it by every term on both sides of the equation.

Problem No. 2 - Solve for x:      The LCD is x. Multiply all terms on both sides of the equation by x.

            8 - 4 = x     x = 4    the ans.

Problem No. 8 - Solve for x:      The LCD is (x - 2), which will be multiplied by each term of the equation. (both sides)

         =   4 - x = -3x + 6

3x - x = 6 - 4    2x = 2    x = 1 the ans. All these problems can be checked by putting the answer for the variable in the original problem, to see if there is a true statement. If there is, that is the answer, if there isn’t, start over and solve the problem again.

Problem No. 14 - Solve for x:      The LCD is 20, so every term of the equation is multiplied by 20.

      =   4(x + 3) + 5x = 8x + 4  By multiplying by 20, (the LCD), the problem will only have whole numbers left in the problem

                   4x + 12 + 5x = 8x + 4        9x + 12 = 8x + 4

                                                              9x - 8x = 4 - 12

                                                                x = -8   the ans.

Problem No. 16 - Solve for x:    The LCD is x(x - 2). This must be multiplied by each term of the equation.

       In the first term the (x - 2) reduces out and in the second term the (x) reduces out, leaving:

       3x - x + 2 = -4x² + 8x   Oh, Wow! a quadratic equation is coming up.      2x + 2 = -4x² + 8x

            4x² + 2x - 8x + 2 = 0

             4x² - 6x + 2 = 0      Divide every term by 2      2x² - 3x + 1 = 0

To solve a quadratic equation, factor the equation first, and then set each factor equal to zero.

              (2x - 1)(x - 1) = 0      2x - 1 = 0   &   x - 1 = 0    {1/2, 1) the ans.

It is a good idea to check both answers.

Problem No. 25 - Solve for x:        The LCD is:   4x(x – 2)

Remember the LCD is all different factors, (in the denominator) used to the greatest number of times shows in any one denominator.

         The first term the 4 and the

(x - 2) reduce out. The second term just the 4 reduces out. The third term just the x reduces out leaving:    6x + x(x - 2) = 16(x - 2)

              6x + x² - 2x = 16x - 32       x² + 6x - 2x - 16x + 32 = 0

                       x² - 12x + 32 = 0      (x - 8)(x - 4) = 0     x = {4, 8}   the ans.

Problem No. 28 - Solve for x:      The LCD is (x - 2).

            2 + 3x - 6 = 3x - 6

   3x - 4 = 3x - 6     3x - 3x = -6 + 4     0 = -2    a false statement, so the answer to this problem is NO SOLUTION. If at any time, one of the solutions, when substituted into the original equation, produced 0, that is an extraneous root and is thrown out.

An extra problem

       When there are only two terms, one on the left of the equal sign and one on the right, it is called a proportion. To solve a proportion, cross multiply as shown below.

     4(x - 1) = 2(26)

       4x - 4 = 52

           4x = 56

            x = 14      the ans.

Another extra problem

               9(6y - 4) = 4(6y + 6)

                               54y - 36 = 24y + 24

                               54y - 24y = 24 + 36

                                        30y = 60

                                            y = 2     the ans.