1. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §6.6 Rational Equations

2. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §6.4 → Complex Rational Expressions  Any QUESTIONS About HomeWork §6.4 → HW-21 6.4 MTH 55

3. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 3 Bruce Mayer, PE Chabot College Mathematics Solving Rational Equations  In previous Lectures, we learned how to simplify expressions. We now learn to solve a new type of equation. A rational equation is an equation that contains one or more rational expressions. Some examples:  We want determine the value(s) for x that make these Equations TRUE

4. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 4 Bruce Mayer, PE Chabot College Mathematics To Solve a Rational Equation 1.List any restrictions that exist. Numbers that make a denominator equal 0 canNOT possibly be solutions. 2.CLEAR the equation of FRACTIONS by multiplying both sides by the LCM of ALL the denominators present 3.Solve the resulting equation using the addition principle, the multiplication principle, and the Principle of Zero Products, as needed. 4.Check the possible solution(s) in the original equation.

5. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION - Because no variable appears in the denominator, no restrictions exist. The LCM of 5, 2, and 4 is 20, so we multiply both sides by 20 Using the multiplication principle to multiply both sides by the LCM. Parentheses are important! Using the distributive law. Be sure to multiply EACH term by the LCM Simplifying and solving for x. If fractions remain, we have either made a mistake or have not used the LCM of ALL the denominators.

6. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 6 Bruce Mayer, PE Chabot College Mathematics Checking Answers  Since a variable expression could represent 0, multiplying both sides of an equation by a variable expression does NOT always produce an Equivalent Equation COULD be Multiplying by Zero and Not Know it  Thus checking each solution in the original equation is essential.

7. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION - Note that x canNOT equal 0. The Denominator LCM is 15x.

8. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Solve  CHECK tentative Solution, x = 5  The Solution x = 5 CHECKS 

9. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION - Note that x canNOT equal 0. The Denom LCM is x  Thus by Zero Products: x = 3 or x = 4

10. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Solve  CHK: For x = 3 For x = 4  Both of these check, so there are two solutions; 3 and 4

11. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION  Note that y canNOT equal 3 or −3. We multiply both sides of the equation by the Denom LCM.

12. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION - Note that x canNOT equal 1 or −1. Multiply both sides of the eqn by the LCM Because of the restriction above, 1 must be rejected as a solution. This equation has NO solution.

13. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: Because the left side of this equation is undefined when x is 0, we state at the outset that x  0.  Next, we multiply both sides of the equation by the LCD, 4x: Multiplying by the LCD to clear fractions

14. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLN cont. Using the distributive law Locating factors equal to 1 Removing factors equal to 1 Using the distributive law

15. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLN cont. This should check since x  0.  CHECK 8

16. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 16 Bruce Mayer, PE Chabot College Mathematics Rational Eqn CAUTION  When solving rational equations, be sure to list any Division-by-Zero restrictions as part of the first step.  Refer to the restriction(s) as you proceed

17. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: To find all restrictions and to assist in finding the LCD, we factor:  Note that to prevent division by zero x  3 and x  −3.  Next multiply by the LCD, (x + 3)(x – 3), and then use the distributive law

18. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: By LCD Multiplication  Remove factors Equal to One and solve the resulting Eqn Keep in Mind any restrictions

19. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLN cont.: Multiply and Collect Similar terms  A check will confirm that 22 is the solution

20. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Eqn with NO Soln To avoid division by zero, exclude from the expression domain 1 and –1, since these values make one or more of the denominators in the equation equal 0. Distributive property Solve. 3 x – 1 = 2 x + 1 – 6 x 2 – 1 = 3 x – 1 2 x + 1 – 6 x 2 – 1 ( x – 1)( x + 1) = 3 x – 1 2 x + 1 – 6 x 2 – 1 ( x – 1)( x + 1) = – 63( x + 1)2( x – 1) = – 63 x + 32 x + 2 = 6 x + 5 = 1 x Multiply each side by the LCD, ( x –1)( x + 1). Multiply. Distributive property Combine terms. Subtract 5.

21. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Eqn with NO Soln Solve. 3 x – 1 = 2 x + 1 – 6 x 2 – 1 Since 1 is not in the domain, it cannot be a solution of the equation. Substituting 1 in the original equation shows why. Check: = 3 x – 1 2 x + 1 – 6 x 2 – 1 = 3 1 – 1 2 1 + 1 – 6 1 2 – 1 = 3 0 2 2 – 6 0 Since division by 0 is undefined, the given equation has no solution, and the solution set is ∅.

22. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Fcn to Eqn  Given Function:  Find all values of a for which On Board  SOLUTION  On Board  By Function Notation:  Thus Need to find all values of a for which f(a) = 4

23. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  Fcn to Eqn  Solve for a:  First note that a  0. To solve for a, multiply both sides of the equation by the LCD, a: Multiplying both sides by a. Parentheses are important. Using the distributive law

24. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Fcn to Eqn  CarryOut Solution  CHECK Simplifying Getting 0 on one side Factoring Using the principle of zero products  STATE: The solutions are 5 and −1. For a = 5 or a = −1, we have f(a) = 4.

25. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 25 Bruce Mayer, PE Chabot College Mathematics Rational Equations and Graphs  One way to visualize the solution to the last example is to make a graph. This can be done by graphing; e.g., Given  Find x such that f(x) = 4

26. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 26 Bruce Mayer, PE Chabot College Mathematics Rational Equations and Graphs  Graph the function, and on the same grid graph y = g(x) = 4  We then inspect the graph for any x-values that are paired with 4. It appears from the graph that f(x) = 4 when x = 5 or x = −1. 4 5

27. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 27 Bruce Mayer, PE Chabot College Mathematics Rational Equations and Graphs  Graphing gives approximate solutions  Although making a graph is not the fastest or most precise method of solving a rational equation, it provides visualization and is useful when problems are too difficult to solve algebraically

28. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 28 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §6.6 Exercise Set 34, 38, 62  Rational Expressions

29. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 29 Bruce Mayer, PE Chabot College Mathematics All Done for Today Remember: can NOT Divide by ZERO

30. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 30 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

31. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 31 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

32. BMayer@ChabotCollege.edu MTH55_Lec-34_sec_6-6_Rational_Equations.ppt 32 Bruce Mayer, PE Chabot College Mathematics