Question

seatwork perform the indicated operations and simplify if possible. 1. $\frac{3x}{x + 1}+\frac{3}{x + 1}$ 4. $\frac{5x - 3}{x^{2}+2x + 1}+\frac{2x + 4}{x^{2}+2x + 1}$ 2. $\frac{2x - 4}{x + 1}-\frac{3}{x + 1}$ 5. $\frac{7x + 5}{x^{2}-2x - 3}-\frac{2x - 1}{x^{2}-2x - 3}$ 3. $\frac{6x - 3}{x + 5}-\frac{2x + 4}{x + 5}$

Explanation:

Step1: Identify common denominators

All pairs of fractions have the same denominator in each problem. When adding or subtracting fractions with the same denominator, we add or subtract the numerators.

Step2: Solve problem 1

For $\frac{3x}{x + 1}+\frac{3}{x + 1}$, we add the numerators: $\frac{3x+3}{x + 1}=\frac{3(x + 1)}{x + 1}=3$.

Step3: Solve problem 2

For $\frac{2x-4}{x + 1}-\frac{3}{x + 1}$, we subtract the numerators: $\frac{2x-4 - 3}{x + 1}=\frac{2x-7}{x + 1}$.

Step4: Solve problem 3

For $\frac{6x-3}{x + 5}-\frac{2x + 4}{x + 5}$, we subtract the numerators: $\frac{6x-3-(2x + 4)}{x + 5}=\frac{6x-3-2x - 4}{x + 5}=\frac{4x-7}{x + 5}$.

Step5: Solve problem 4

For $\frac{5x-3}{x^{2}+2x + 1}+\frac{2x + 4}{x^{2}+2x + 1}$, first factor $x^{2}+2x + 1=(x + 1)^{2}$. Then add the numerators: $\frac{5x-3+2x + 4}{(x + 1)^{2}}=\frac{7x+1}{(x + 1)^{2}}$.

Step6: Solve problem 5

For $\frac{7x + 5}{x^{2}-2x-3}-\frac{2x-1}{x^{2}-2x-3}$, first factor $x^{2}-2x-3=(x-3)(x + 1)$. Then subtract the numerators: $\frac{7x + 5-(2x-1)}{(x-3)(x + 1)}=\frac{7x + 5-2x + 1}{(x-3)(x + 1)}=\frac{5x+6}{(x-3)(x + 1)}$.

Answer:

1. $3$
2. $\frac{2x-7}{x + 1}$
3. $\frac{4x-7}{x + 5}$
4. $\frac{7x+1}{(x + 1)^{2}}$
5. $\frac{5x+6}{(x-3)(x + 1)}$