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What’s Ahead: Today: Lecture on Section 6.5 ( Solving equations containing rational expressions) Next class: Lecture on Section 6.6 ( Solving applied problems with rational expressions) Required paper worksheet for this assignment MUST be turned in at the start of class when the homework is due or your online score will be reduced to ZERO. Next two class sessions after that: – Review for Test 3 on all of Ch. 6 – Test 3 (You should be working on the practice test as soon as you finish section 6.6 – don’t wait till the last minute!)

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Now please CLOSE YOUR LAPTOPS and turn off and put away your cell phones. Sample Problems Page Link (Dr. Bruce Johnston )

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Section 6.5 Solving Equations Containing Rational Expressions

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To review: Recall the difference between “Factor” and “Solve” problems from the previous chapter: Factor x 3 + 7x 2 + 6x vs. Solve x 3 + 7x 2 + 6x = 0 Also review this problem: Solve x 3 – 5 x 2 = 4x – 20 3 3 In this new section, we’re going to move on to solving equations involving rational expressions instead of just simplifying or adding/subtracting/multiplying/dividing rational expressions.

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Equations with Rational Expressions First note that an equation contains an equal sign and an expression does not. To solve EQUATIONS containing rational expressions, clear the fractions by multiplying only the numerators on both sides of the equation by the LCD of all the fractions. Then solve as in previous sections, with the additional step of making sure any solutions you find are in the domain of the original rational expressions.

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Example Solve the following rational equation. So 10 appears to be the solution

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Two steps to checking your answer: Step 1. Is it in the DOMAIN of the rational expressions in the original problem? Tentative solution: x = 10 Domain: {x | x ≠ 0} Therefore 10 is a valid possible solution. Step 2. Does the answer check in the original equation?

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true So the solution is x = 10. CHECK: Substitute 10 for x into the equation.

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Example: Solve x + 5 = 2. x + 3 x + 3 Solution: 1. Multiply both sides by the LCD (x + 3). This gives x + 5 = 2 2. Now solve this equation: Answer: x = -3 3. NOW CHECK in original equation: -3 + 5 = 2 -3 + 3 0 This is undefined, so the answer is ø (no solution)

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Example (cont.) THIS IS WHY YOU MUST ALWAYS CHECK YOUR ANSWER ON THESE PROBLEMS WITH X’S IN THE DENOMINATOR! IF YOU ANSWERED “x = -3” TO THIS PROBLEM, YOU’D GET IT WRONG (and no partial credit, either!) Why? Because an answer must be in the DOMAIN of the rational expression to be a valid solution.

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Example Solve the following rational equation.

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Example (cont.) true Substitute the value for x into the original equation, to check the solution. So the solution is ALWAYS CHECK!!

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Example Solve the following rational equation.

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Potential answer: x = -7/5 Two steps to checking your answer: Step 1: Is the potential answer in the DOMAIN of the rational expression? Domain: All real numbers EXCEPT those that make the denominator zero. The LCD from previous problem was 3(x+2)(x+5), so the domain is all real numbers except -2 and -5. So -7/5 passes the first step of the check.

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true Step 2 of the answer check: Substitute the value for x into the original equation, to check the solution. Potential answer: x = -7/5 So the solution is Original equation:

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Example Solve the following rational equation.

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Example (cont.) Substitute the value for x into the original equation, to check the solution. true So the solution is x = 3. ALWAYS CHECK!! And also be sure to check: Is x=3 in the domain of the original equation? Domain: x ≠ 1, -1

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Example Solve the following rational equation.

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Example Solve the following rational equation.

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Example (cont.) Substitute the value for a into the original equation, to check the solution. Since substituting the suggested value of a into the equation produced undefined expressions, the solution is . A proposed solution that is rejected is referred to as an extraneous solution. ALWAYS CHECK!!

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Example (cont.) THIS IS WHY YOU MUST ALWAYS CHECK YOUR ANSWER ON THESE PROBLEMS WITH X’S IN THE DENOMINATOR! IF YOU ANSWERED “X = 3” TO THIS PROBLEM, YOU’D GET IT WRONG (and no partial credit, either!)

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Example: Solve 2x + 1 = 1. 2x – 1 x 2x – 1 Solution: 1. Multiply both sides by the LCD x(2x – 1). This gives 2x(x) + 1(2x – 1) = 1(x) 2. Now simplify this equation: 2x 2 + 2x – 1 = x 2x 2 + x – 1 = 0 3. Solve either by factoring or by using the quadratic formula. Answers: x = -1, x = 1/2

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Example (continued): NOW CHECK BOTH ANSWERS in original equation: Note that when you plug in x = 1, everything checks out OK ( you get -1/3 = -1/3). BUT when you plug in x = 1/2, you get zeroes in some denominators, so this number is NOT a solution. (it gives 1 + 1 = 1 ) 0 ½ 0 So the final answer is just x = -1 (if you write x = -1, 1/2, you’ll get ZERO credit.)

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Reminder: This homework assignment on section 6.5 is due at the start of next class period. NOTE: The 6.6 online homework due the session after next has a required offline (paper) worksheet. Pick it up today if you want to start working on it early (or you can also print it off the assignment web site.)

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You may now OPEN your LAPTOPS and begin working on the homework assignment.

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