Introduction Your previous experiences have helped you understand how to represent situations using algebraic equations, and how solving those equations can reveal something specific about those situations. Most of these equations have involved integer coefficients. However, coefficients can also be expressed as a ratio. A ratio compares two quantities, and is often written as a fraction. Numbers expressed as a ratio are called rational numbers : Creating and Solving Rational Equations

Introduction, continued More formally, a rational number is any number that can be written as, where m and n are integers and n ≠ 0. In this lesson, we will explore ways to solve rational equations, and to assess and validate the resulting solutions : Creating and Solving Rational Equations

Key Concepts A rational equation is an algebraic equation that includes at least one term that is expressed as a ratio. A rational equation defines a specific relationship between numbers and variables; this relationship can be analyzed to find a solution to the equation. Finding the solution to a rational equation may be accomplished by simplifying the rational terms. To simplify rational terms, convert the rational equation into an equivalent equation with integer coefficients : Creating and Solving Rational Equations

Key Concepts, continued Finding the solution often requires using a common denominator—a quantity that is a shared multiple of the denominators of two or more fractions—to eliminate any rational terms. Recall that the least common denominator is the least common multiple of the denominators of the fractions in the expression. Like other algebraic equations, a rational equation may have no solutions : Creating and Solving Rational Equations

Key Concepts, continued An extraneous solution can occur when the solution process results in a value that satisfies the original equation, but creates an invalid equation by creating a denominator that is equal to 0. An extraneous solution can also occur when a solution results in a value that is irrelevant in the context of the original equation : Creating and Solving Rational Equations

Key Concepts, continued Recall that a proportion is a statement of equality between two ratios. Techniques for solving proportions can often be used to simplify rational equations. The process of solving rational equations can be applied to real-world examples in which a specific task that may be done alone is to be completed cooperatively : Creating and Solving Rational Equations

Common Errors/Misconceptions multiplying only some terms of the rational equation when eliminating rational numbers incorrectly or prematurely reducing rational numbers eliminating terms instead of factors neglecting to check for extraneous solutions : Creating and Solving Rational Equations

Guided Practice Example 2 Solve the rational equation for t : Creating and Solving Rational Equations

Guided Practice: Example 2, continued 1.Determine the least common denominator. The denominators in the equation are t, 12, and 3t. The smallest multiple having all three of these values as a factor is 12t; therefore, the least common denominator is 12t : Creating and Solving Rational Equations

Guided Practice: Example 2, continued 2.Multiply each term of the equation by the common denominator and simplify. Multiply each term by the LCD, 12t. Simplify each fraction. (12)5 – (t)3 = (4)9 Continue to simplify. 60 – 3t = : Creating and Solving Rational Equations

Guided Practice: Example 2, continued The result, 60 – 3t = 36, is an algebraic equation without any remaining rational numbers that can be solved using methods you have already learned : Creating and Solving Rational Equations

Guided Practice: Example 2, continued 3.Solve the resulting equation. 60 – 3t = 36 Simplified equation from the previous step 60 = t Add 3t to both sides. 24 = 3t Subtract 36 from both sides. t = 8 Divide both sides by : Creating and Solving Rational Equations

Guided Practice: Example 2, continued 4.Verify the solution(s). While it appears that we have a solution, t = 8, we have to return to the original equation to confirm it. Because there are variables in the denominators of the first and last terms, we must determine what value(s) would make the denominator equal 0, and thereby invalidate the equation (since division by 0 yields an undefined result and is not acceptable) : Creating and Solving Rational Equations

Guided Practice: Example 2, continued The expression is undefined when t = 0; therefore, t ≠ 0. Similarly, the expression is also undefined when t = 0; once again, t ≠ 0. Our solution, t = 8, does not violate this condition; therefore, it is an acceptable result. For the equation, t = : Creating and Solving Rational Equations ✔

Guided Practice: Example 2, continued : Creating and Solving Rational Equations

Guided Practice Example 3 Solve the rational equation for y : Creating and Solving Rational Equations

Guided Practice: Example 3, continued 1.Determine the least common denominator. The denominators are y, y + 2, and 3. The smallest multiple having all three of these values as a factor is 3y(y + 2), the irreducible product of all three denominators. The least common denominator is 3y(y + 2) : Creating and Solving Rational Equations

Guided Practice: Example 3, continued 2.Multiply each term by the common denominator : Creating and Solving Rational Equations

Guided Practice: Example 3, continued 3.Simplify the resulting equation. Start by rewriting each occurrence of the common denominator in the equation as a ratio: : Creating and Solving Rational Equations Rewritten equation Common denominator rewritten as a ratio

Guided Practice: Example 3, continued 3(y + 2) 2 – 3y 1 = Simplify the ratios. y(y + 2) 1 6y + 12 – 3y = y 2 + 2y Carry out the multiplication. 3y + 12 = y 2 + 2y Subtract 3y from 6y : Creating and Solving Rational Equations

Guided Practice: Example 3, continued The algebraic equation that results this time, 3y + 12 = y 2 + 2y, still has only integer coefficients. However, it is no longer linear—this is a quadratic equation because y is raised to a power of 2. We will need to either factor or use the quadratic formula to determine the value of y : Creating and Solving Rational Equations

Guided Practice: Example 3, continued 4.Solve the quadratic equation by factoring or using the quadratic formula. Start by setting the equation equal to 0 so that the equation is in the standard form of a quadratic equation. Then, factor the quadratic : Creating and Solving Rational Equations

Guided Practice: Example 3, continued 3y + 12 = y 2 + 2y Equation found in the previous step 12 = y 2 + (– y)Subtract 3y from both sides. 0 = y 2 – y – 12 Subtract 12 from both sides. 0 = (y – 4)(y + 3) Factor the resulting quadratic. y = 4 or y = – : Creating and Solving Rational Equations

Guided Practice: Example 3, continued 5.Verify the solution(s). Now that we have potential solutions, y = 4 or y = –3, we must return to the original equation. We must determine what value(s) would make the denominators equal 0, and thereby invalidate the equation : Creating and Solving Rational Equations

Guided Practice: Example 3, continued The expression is undefined when y = 0; therefore, y ≠ 0. The expression is also undefined when y = –2; therefore, y ≠ –2. There is no value for y that invalidates the last term,, because the denominator does not contain a variable : Creating and Solving Rational Equations

Guided Practice: Example 3, continued Our solutions, y = 4 or y = –3, do not violate the condition that the value of y cannot make the denominators equal 0; therefore, they are acceptable results. If, then y = 4 or y = – : Creating and Solving Rational Equations ✔

Guided Practice: Example 3, continued : Creating and Solving Rational Equations