Recomposition Math Example 2

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Example 2

medium
Partial fractions gave 5x+1(x+1)(x−2)=43(x+1)+113(x−2)\frac{5x+1}{(x+1)(x-2)} = \frac{4}{3(x+1)}+\frac{11}{3(x-2)}. Recompose to verify by adding the fractions.

Solution

  1. 1
    Common denominator is 3(x+1)(x−2)3(x+1)(x-2).
  2. 2
    43(x+1)=4(x−2)3(x+1)(x−2)\frac{4}{3(x+1)} = \frac{4(x-2)}{3(x+1)(x-2)} and 113(x−2)=11(x+1)3(x+1)(x−2)\frac{11}{3(x-2)} = \frac{11(x+1)}{3(x+1)(x-2)}.
  3. 3
    Sum: 4(x−2)+11(x+1)3(x+1)(x−2)=4x−8+11x+113(x+1)(x−2)=15x+33(x+1)(x−2)=5x+1(x+1)(x−2)\frac{4(x-2)+11(x+1)}{3(x+1)(x-2)} = \frac{4x-8+11x+11}{3(x+1)(x-2)} = \frac{15x+3}{3(x+1)(x-2)} = \frac{5x+1}{(x+1)(x-2)}. Confirmed.

Answer

43(x+1)+113(x−2)=5x+1(x+1)(x−2)  ✓\frac{4}{3(x+1)}+\frac{11}{3(x-2)} = \frac{5x+1}{(x+1)(x-2)}\;\checkmark
Recomposition reassembles the decomposed pieces and verifies them against the original. Adding partial fractions back together confirms the decomposition is correct.

About Recomposition

Recomposition is the process of combining simpler parts, sub-results, or solved sub-problems back together to form a complete solution or to understand the whole structure from its pieces.

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More Recomposition Examples

Example 1 easy

After decomposing [formula], recompose to solve [formula].

Example 3 easy

You find [formula] and [formula] from solving a system. Recompose by substituting back into both equ

Example 4 medium

Given the identity [formula], recompose: use it to evaluate [formula].

← All Recomposition Examples Recomposition Concept

Related Concepts

Decomposition