Solving Rational Equations

7 Key excerpts on "Solving Rational Equations"

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  eBook - PDF

  Elementary Algebra 2e

  • Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis(Authors)
  • 2020(Publication Date)
  • Openstax

    (Publisher)

  We multiplied both sides by the LCD. Then we distributed. We simplified—and then we had an equation with no fractions. Finally, we solved that equation. We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator we must be careful that we don’t end up with a solution that would make a denominator equal to zero. So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard. An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution. Extraneous Solution to a Rational Equation An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined. We note any possible extraneous solutions, c, by writing x ≠ c next to the equation. EXAMPLE 8.59 HOW TO SOLVE EQUATIONS WITH RATIONAL EXPRESSIONS Solve: 1 x + 1 3 = 5 6 . Solution 970 Chapter 8 Rational Expressions and Equations This OpenStax book is available for free at http://cnx.org/content/col31130/1.4 TRY IT : : 8.117 Solve: 1 y + 2 3 = 1 5 . TRY IT : : 8.118 Solve: 2 3 + 1 5 = 1 x . The steps of this method are shown below. We always start by noting the values that would cause any denominators to be zero. EXAMPLE 8.60 Solve: 1 − 5 y = − 6 y 2 . HOW TO : : SOLVE EQUATIONS WITH RATIONAL EXPRESSIONS. Note any value of the variable that would make any denominator zero. Find the least common denominator of all denominators in the equation. Clear the fractions by multiplying both sides of the equation by the LCD.

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  Beginning and Intermediate Algebra

  A Guided Approach

  • Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  EXAMPLE 1 Solution a SELF CHECK 1 2 EXAMPLE 2 Solution Copyright 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6.5 Solving Equations That Contain Rational Expressions 407 x 1 3 5 4 Simplify. x 5 1 Subtract 3 from both sides. Because both sides were multiplied by an expression containing a variable, we must check to see if the apparent solution is a value that must be excluded. If we replace x with 1 in the original equation, both denominators will become 0. Therefore, 1 is not a solution. Such false solutions are often called extraneous solutions . Because 1 does not satisfy the original equation, there is no solution.

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  Introductory Algebra

  Concepts with Applications

  • Charles P. McKeague(Author)
  • 2013(Publication Date)
  • XYZ Textbooks

    (Publisher)

  OBJECTIVES 480 KEY WORDS Chapter 6 Rational Expressions and Equations rational equation A Solve equations that contain rational expressions 6.4 Equations Involving Rational Expressions For 520 days, 6 men will simulate a trip to Mars and back. The men will be locked inside a large facility that was built to look like the interior of a space shuttle. Isolated from the outside world, the men will study the effects of the mission on their own bodies. The mission includes a trip to the planet for x number of days, y number of days to perform mock experiments on the planet, and then z number of days to fly back to Earth. We can set up the following rational expressions to represent the days in each stage of the mission: Trip to Mars Trip to Earth x ___ 520 + y ___ 520 + z ___ 520 =Total days of mission ↑ Experiments Then if given the value for two of the variables, we can determine the value of the third variable using a rational equation. Equations that involve rational expressions are the focus of this section. A Solving Rational Equations The first step in solving an equation that contains one or more rational expressions is to find the LCD for all denominators in the equation. Once the LCD has been found, we multiply both sides of the equation by it. The resulting equation should be equivalent to the original one (unless we inadvertently multiplied by zero) and free from any denominators except the number 1. EXAMPLE 1 Solve x __ 3 + 5 __ 2 = 1 __ 2 . Solution The LCD for 3 and 2 is 6. If we multiply both sides by 6, we have 6  x __ 3 + 5 __ 2  = 6  1 __ 2  Multiply both sides by 6. 6  x __ 3  + 6  5 __ 2  = 6  1 __ 2  Distributive property 2x + 15 = 3 2x = −12 x = −6 We can check our solution by replacing x with −6 in the original equation. − 6 __ 3 + 5 __ 2 ≟ 1 __ 2 1 __ 2 = 1 __ 2 The solution is −6. Answer 1. − 4 _ 5 Practice Problems 1. Solve x __ 2 + 3 __ 5 = 1 __ 5 . Image © NASA JPL

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  Intermediate Algebra

  • Jerome Kaufmann, Karen Schwitters, , , Jerome Kaufmann, Karen Schwitters(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Keep in mind that listing the restrictions at the beginning of a problem does not replace check-ing the potential solutions. In Example 4, the answer 11 needs to be checked in the original equation. Solve a a 2 2 1 2 3 5 2 a 2 2 . Solution a a 2 2 1 2 3 5 2 a 2 2 , a ? 2 3( a 2 2) a a a 2 2 1 2 3 b 5 3( a 2 2) a 2 a 2 2 b Multiply both sides by 3( a 2 2) 3( a 2 2) a a a 2 2 b 1 3( a 2 2) a 2 3 b 5 3( a 2 2) a 2 a 2 2 b 3( a ) 1 2( a 2 2) 5 3(2) 3 a 1 2 a 2 4 5 6 5 a 5 10 a 5 2 Because our initial restriction was a ? 2 , we conclude that this equation has no solution. Thus the solution set is [ . Solving Proportions A ratio is the comparison of two numbers by division. We often use the fractional form to ex-press ratios. For example, we can write the ratio of a to b as a b . A statement of equality between two ratios is called a proportion . Thus if a b and c d are two equal ratios, we can form the propor-tion a b 5 c d ( b ? 0 and d ? 0 ). We deduce an important property of proportions as follows: a b 5 c d , b ? 0 and d ? 0 bd a a b b 5 bd a c d b Multiply both sides by bd ad 5 bc Cross-Multiplication Property of Proportions If a b 5 c d ( b ? 0 and d ? 0 ), then ad 5 bc . EXAMPLE 5 Classroom Example Solve x x 1 3 1 3 2 5 2 3 x 1 3 . Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Chapter 4 • Rational Expressions 220 We can treat some fractional equations as proportions and solve them by using the cross-multiplication idea, as in the next examples.

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  Elementary Algebra

  • Mark D. Turner, Charles P. McKeague(Authors)
  • 2017(Publication Date)
  • XYZ Textbooks

    (Publisher)

  Solve number problems involving rational expressions. 2. Solve motion problems involving rational expressions. 3. Solve work problems involving rational expressions. In this section we will solve some word problems whose equations involve rational expressions. Like the other word problems we have encountered, the more you work with them, the easier they become. Number Problems One number is twice another. The sum of their reciprocals is 9 _ 2 . Find the two numbers. SOLUTION Let x represent the smaller number. The larger then must be 2 x . Their reciprocals are 1 _ x and 1 __ 2 x , respectively. An equation that describes the situation is: 1 __ x + 1 __ 2 x = 9 __ 2 We can multiply both sides by the LCD of 2 x and then solve the resulting equation: 2 x  1 __ x  + 2 x  1 __ 2 x  = 2 x  9 __ 2  2 + 1 = 9 x 3 = 9 x x = 3 __ 9 = 1 __ 3 The smaller number is 1 _ 3 . The other number is twice as large, or 2 _ 3 . If we add their reciprocals, we have: 3 __ 1 + 3 __ 2 = 6 __ 2 + 3 __ 2 = 9 __ 2 The solutions check with the original problem. Motion Problems Recall from Section 7.1 that if an object travels at a constant rate r for a specified time t , then the distance traveled is given by the rate equation Distance = Rate ⋅ Time d = rt If we know the distance traveled and the rate, then we can find the time by dividing the distance by the rate: Time = Distance _______ Rate t = d __ r EXAMPLE 1 VIDEO EXAMPLES SECTION 7.7 Applications 500 Chapter 7 Rational Expressions The next two examples use this version of the rate equation to solve problems involving motion. A boat travels 30 miles up a river in the same amount of time it takes to travel 50 miles down the same river. If the current is 5 miles per hour, what is the speed of the boat in still water? SOLUTION The easiest way to work a problem like this is with a table. The top row of the table is labeled with d for distance, r for rate, and t for time.

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  Years 9 - 10 Maths For Students

  • (Author)
  • 2015(Publication Date)
  • For Dummies

    (Publisher)

  Proportions are very nice to work with because of their unique properties of reducing and changing into non‐fractional equations. Many equations involving fractions must be dealt with in that fractional form, but other equations are easily changed into proportions. When possible, you want to take advantage of the situations where transformations can be done.

  For example, solve the following equation for x:

  You could solve the problem by multiplying each fraction by the least common factor of all the fractions: 24. Another option is to find a common denominator for the two fractions on the left and subtract them, and then find a common denominator for the two fractions on the right and add them. Your result is a proportion:

  The proportion can be reduced by dividing by 2 horizontally:

  Now cross‐multiply and simplify the products:

  (−3x + 3) ⋅ 4 = 3 ⋅ (13x − 13)

   −12x + 12 = 39x − 39

  Add 12x to each side, and then add 39 to each side:

  The last step consists of just dividing each side by 51 to get 1 = x.

  Solving for Variables in Formulas

  A formula is an equation that represents a relationship between some structures or quantities or other entities. It’s a rule that uses mathematical computations and can be counted on to be accurate each time you use it when applied correctly. The following are some of the more commonly used formulas that contain only variables raised to the first power.

  • : The area of a triangle involves base and height.
  • I = Prt: The interest earned uses principal, rate and time.
  • C = 2πr: Circumference is twice times the radius.
  • : Degrees Fahrenheit uses degrees Celsius.
  • P = R − C: Profit is based on revenue and cost.

  When you use a formula to find the indicated variable (the one on the left of the equal sign), you just put the numbers in, and out pops the answer. Sometimes, though, you’re looking for one of the other variables in the equation and end up solving for that variable over and over.

  For example, let’s say that you’re helping your mum plan a circular rose garden in your backyard. You find edging on sale and can buy a 20‐metre roll of edging, a 36‐metre roll, a 40‐metre roll or a 48‐metre roll. you’re going to use every bit of the edging and let the length of the roll dictate how large the garden will be. If you want to know the radius of the garden based on the length of the roll of edging, you use the formula for circumference and solve the following four equations:

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  Years 9 - 10 Maths For Students

  • (Author)
  • 2015(Publication Date)
  • For Dummies

    (Publisher)

  Transforming fractional equations into proportions Proportions are very nice to work with because of their unique properties of reducing and changing into non‐fractional equations. Many equations involving fractions must be dealt with in that fractional form, but other equations are easily changed into proportions. When possible, you want to take advantage of the situations where transformations can be done. 240 Part III: Solving Algebraic Equations For example, solve the following equation for x: 2 3 5 1 6 3 1 2 9 8 x x x x + − + = − + − You could solve the problem by multiplying each fraction by the least common factor of all the fractions: 24. Another option is to find a common denominator for the two fractions on the left and subtract them, and then find a common denominator for the two fractions on the right and add them. Your result is a proportion: + ⋅ − + = − ⋅ + − + − + = − + − + − − = − + − − + = − x x x x x x x x x x x x x x 2( 2) 2 3 5 1 6 4(3 1) 4 2 9 8 2( 2) (5 1) 6 4(3 1) ( 9) 8 2 4 5 1 6 12 4 9 8 3 3 6 13 13 8 The proportion can be reduced by dividing by 2 horizontally: 3 3 6 13 13 8 3 4 x x − + = − Now cross‐multiply and simplify the products: (−3x + 3) ⋅ 4 = 3 ⋅ (13x − 13) −12x + 12 = 39x − 39 Add 12x to each side, and then add 39 to each side: x x x x x x − + = − + + = − + + = 12 12 39 39 12 12 12 51 39 39 39 51 51 The last step consists of just dividing each side by 51 to get 1 = x. 241 Chapter 10: Establishing the Ground Rules and Solving Linear Equations Solving for Variables in Formulas A formula is an equation that represents a relationship between some structures or quantities or other entities. It’s a rule that uses mathematical computations and can be counted on to be accurate each time you use it when applied correctly. The following are some of the more commonly used formulas that contain only variables raised to the first power. 6 = A A bh 1 2 : The area of a triangle involves base and height.

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