Math Problem Statement

Solution

The given equation to solve for uu is:

1u5+5u+3=2u22u15\frac{1}{u - 5} + \frac{5}{u + 3} = \frac{2}{u^2 - 2u - 15}

Step 1: Factor the quadratic expression in the denominator

First, factor the quadratic expression u22u15u^2 - 2u - 15:

u22u15=(u5)(u+3)u^2 - 2u - 15 = (u - 5)(u + 3)

So the equation becomes:

1u5+5u+3=2(u5)(u+3)\frac{1}{u - 5} + \frac{5}{u + 3} = \frac{2}{(u - 5)(u + 3)}

Step 2: Multiply through by the common denominator

The common denominator for all terms is (u5)(u+3)(u - 5)(u + 3). Multiply every term by this common denominator to eliminate the fractions:

(u5)(u+3)1u5+(u5)(u+3)5u+3=(u5)(u+3)2(u5)(u+3)(u - 5)(u + 3) \cdot \frac{1}{u - 5} + (u - 5)(u + 3) \cdot \frac{5}{u + 3} = (u - 5)(u + 3) \cdot \frac{2}{(u - 5)(u + 3)}

Simplifying, we get:

(u+3)+5(u5)=2(u + 3) + 5(u - 5) = 2

Step 3: Expand and simplify the equation

Expand and combine like terms:

u+3+5u25=2u + 3 + 5u - 25 = 2

6u22=26u - 22 = 2

Step 4: Solve for uu

Add 22 to both sides:

6u=246u = 24

Divide by 6:

u=4u = 4

Step 5: Check the solution

Substitute u=4u = 4 back into the original equation to verify that it does not cause any denominators to be zero:

  • u5=45=1u - 5 = 4 - 5 = -1 (not zero)
  • u+3=4+3=7u + 3 = 4 + 3 = 7 (not zero)
  • u22u15=422(4)15=16815=7u^2 - 2u - 15 = 4^2 - 2(4) - 15 = 16 - 8 - 15 = -7 (not zero)

Since none of the denominators are zero, u=4u = 4 is a valid solution.

Final Answer:

The solution to the equation is u=4u = 4.

Would you like to ask any further questions, or do you need more details on this problem?

Related Questions:

  1. What are the steps to factor a quadratic expression?
  2. How do you identify a common denominator in a rational equation?
  3. Why is it important to check solutions for potential undefined expressions?
  4. How can you simplify rational expressions before solving an equation?
  5. What alternative methods exist to solve rational equations?
  6. Can this problem have multiple solutions, and why or why not?
  7. How would the solution process change if the quadratic had no real factors?
  8. What is the significance of extraneous solutions in rational equations?

Tip:

Always check your solution by substituting it back into the original equation to ensure that it does not make any denominators zero or introduce any undefined expressions.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rational Equations
Quadratic Equations

Formulas

Quadratic equation factorization

Theorems

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Suitable Grade Level

Grades 10-12