Adding with Linear Denominators

Simplify: $\frac{2}{x+3} + \frac{5}{x-2}$

Identify the denominators: Denominator 1: $(x+3)$, Denominator 2: $(x-2)$ = Both are linear, no common factors

Find the LCD: LCD = $(x+3)(x-2)$ = Multiply the distinct factors

Rewrite first fraction: $\frac{2}{x+3} \cdot \frac{x-2}{x-2} = \frac{2(x-2)}{(x+3)(x-2)}$ = $\frac{2x-4}{(x+3)(x-2)}$

Rewrite second fraction: $\frac{5}{x-2} \cdot \frac{x+3}{x+3} = \frac{5(x+3)}{(x+3)(x-2)}$ = $\frac{5x+15}{(x+3)(x-2)}$

Add the numerators: $\frac{2x-4+5x+15}{(x+3)(x-2)}$ = $\frac{7x+11}{(x+3)(x-2)}$

Answer: $\frac{7x+11}{(x+3)(x-2)}$ or $\frac{7x+11}{x^2+x-6}$

Adding with a Factorable Denominator

Simplify: $\frac{3}{x^2-4} + \frac{2}{x+2}$

Factor all denominators: $x^2-4 = (x+2)(x-2)$ = First denominator factors as difference of squares

Find the LCD: LCD = $(x+2)(x-2)$ = $(x+2)$ is already in the factored form

Rewrite fractions with LCD: $\frac{3}{(x+2)(x-2)}$ stays the same; $\frac{2}{x+2} \cdot \frac{x-2}{x-2} = \frac{2(x-2)}{(x+2)(x-2)}$ = Both now have the same denominator

Add the numerators: $\frac{3 + 2(x-2)}{(x+2)(x-2)} = \frac{3 + 2x - 4}{(x+2)(x-2)}$ = $\frac{2x-1}{(x+2)(x-2)}$

Check for simplification: $2x-1$ does not factor with $(x+2)$ or $(x-2)$ = Already in simplest form

Answer: $\frac{2x-1}{(x+2)(x-2)}$ or $\frac{2x-1}{x^2-4}$

Adding Three Rational Expressions

Simplify: $\frac{1}{x} + \frac{2}{x+1} + \frac{3}{x-1}$

Identify all denominators: $x$, $(x+1)$, $(x-1)$ = Three distinct linear factors

Find the LCD: LCD = $x(x+1)(x-1)$ = Product of all factors

Rewrite each fraction: $\frac{(x+1)(x-1)}{x(x+1)(x-1)} + \frac{2x(x-1)}{x(x+1)(x-1)} + \frac{3x(x+1)}{x(x+1)(x-1)}$ = All fractions now have LCD

Expand numerators: $(x^2-1) + (2x^2-2x) + (3x^2+3x)$ = $6x^2 + x - 1$

Write final answer: $\frac{6x^2+x-1}{x(x+1)(x-1)}$ = Check if numerator factors

Answer: $\frac{6x^2+x-1}{x(x+1)(x-1)}$ or $\frac{6x^2+x-1}{x^3-x}$

Mistake: Forgetting to factor denominators first

Why: Without factoring, you might miss common factors and create an LCD that is larger than necessary, making the problem harder.

Correct: Always factor each denominator completely before finding the LCD. For example, $x^2-9 = (x+3)(x-3)$.

Mistake: Adding denominators instead of finding LCD

Why: Unlike numerators, denominators are not added together. The LCD must contain all factors from each denominator.

Correct: $\frac{1}{x+2} + \frac{1}{x+3} \neq \frac{2}{2x+5}$. The LCD is $(x+2)(x+3)$.

Mistake: Forgetting to multiply both numerator and denominator

Why: When converting to the LCD, you must multiply by a form of 1 (same expression over itself) to keep the value unchanged.

Correct: $\frac{2}{x+1} = \frac{2(x-1)}{(x+1)(x-1)}$, not $\frac{2}{(x+1)(x-1)}$.

Mistake: Not simplifying the final answer

Why: After adding, the resulting numerator might share a common factor with the denominator.

Correct: Always check if the numerator can be factored and if any factors cancel with the denominator.

Parallel Resistors in Electronics

When resistors are connected in parallel, the total resistance is found using the formula with unlike denominators.

For two resistors of $R_1 = x$ ohms and $R_2 = x+2$ ohms in parallel: $\frac{1}{R_{total}} = \frac{1}{x} + \frac{1}{x+2} = \frac{2x+2}{x(x+2)}$

Combined Work Rate Problems

When two people work together, their combined rate involves adding fractions with different denominators.

If Worker A completes a job in $x$ hours and Worker B in $(x+2)$ hours, their combined rate is $\frac{1}{x} + \frac{1}{x+2} = \frac{2x+2}{x(x+2)}$ jobs per hour.

Q: How do I find the LCD of two rational expressions?

A: Factor each denominator completely. The LCD contains each factor raised to the highest power it appears in any denominator. For example, for $\frac{1}{x^2(x+1)}$ and $\frac{1}{x(x+1)^2}$, the LCD is $x^2(x+1)^2$.

Q: What if the denominators share a common factor?

A: If denominators share factors, include each shared factor only once in the LCD. For example, $(x+2)(x-1)$ and $(x+2)(x+3)$ share $(x+2)$, so the LCD is $(x+2)(x-1)(x+3)$.

Q: Do I need to expand the denominator in my final answer?

A: Either form is acceptable. Factored form $(x+2)(x-3)$ is often preferred because it makes domain restrictions clearer and simplification easier. Expanded form $x^2-x-6$ may be requested in some contexts.