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Algebraic Fractions and Rational Equations

Algebraic Fractions and Rational Equations. In this discussion, we will look at examples of simplifying Algebraic Fractions using the 4 rules of fractions. +. -. ×. ÷. Addition. Simplify. +. Addition. Simplify. +. Subtraction. Simplify. -. Subtraction. Simplify. 4. -.

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Algebraic Fractions and Rational Equations

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  1. Algebraic Fractions and Rational Equations

  2. In this discussion, we will look at examples of simplifying Algebraic Fractions using the 4 rules of fractions. + - × ÷

  3. Addition Simplify +

  4. Addition Simplify +

  5. Subtraction Simplify -

  6. Subtraction Simplify 4 -

  7. Multiplication Simplify ×

  8. Multiplication Simplify Factoring (2x-6) 3 ×

  9. Division Simplify 2 2 ÷

  10. Equations Once you know how to simplify algebraic fractions, you Can solve equations containing them For example Click an equation to see it solved

  11. Solve Back

  12. Solve Back

  13. and are rational equations. To solve a rational equation: Rational Equation The equation in the previous example is called a rational equation. 1. Find the LCM of the denominators. 2. Clear denominators by multiplying both sides of theequation by the LCM. 3. Solve the resulting polynomial equation. 4. Check the solutions.

  14. Examples: 1. Solve: . (0) (0) (0) 2. Solve: . Examples: Solve 1= x + 1 Multiply by LCM = (x – 3). x = 0 Solve for x. Check. True. Simplify. LCM = x(x – 1). Find the LCM. Multiply by LCM. x – 1 = 2x Simplify. x = –1 Solve.

  15. Example:Solve: . Example: Solve After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero. In this case, the value is not a solution of the rational equation. It is critical to check all solutions. Since x2 – 1 = (x – 1)(x +1), LCM = (x – 1)(x + 1). 3x + 1 = x – 1 2x = –2 x = –1 Since –1makes both denominators zero, the rational equation has no solutions. Check.

  16. Example: Solve: . Example: Solve x2 – 8x + 15 = (x – 3)(x – 5) Factor. The LCM is (x – 3)(x – 5). Original Equation. x(x – 5) = –6 Polynomial Equation. x2 – 5x + 6 = 0 Simplify. (x – 2)(x – 3) = 0 Factor. Check. x = 2 is a solution. x = 2 or x = 3 Check. x =3 is not a solution since both sides would be undefined.

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