6. $$ \frac{x+2}{x-1} \div \frac{x+4}{x^{2}+4 x-5} $$ a. $$ \frac{(x+2)(x+5)}{x+4}, x
eq-5,-4 $$ c. $$ \frac{(x+2)(x+4)}{(x-1)^{2}(x+5)}, x
eq 1,-5,-4 $$ b. $$ \frac{(x+2)(x+4)}{(x-1)^{2}(x+5)}, x
eq 1,-5 $$ d. $$ \frac{(x+2)(x+5)}{x+4}, x
eq 1,-4,-5 $$ 7. $$ \frac{x^{2}-4}{x-8} \div(x-2) $$ a. $$ \frac{(x-2)(x+2)}{x-8} $$ c. $$ \frac{x-2}{x-8} $$ b. $$ \frac{x-8}{x+2} $$ d. $$ \frac{x+2}{x-8} $$ 8. $$ \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 $$ a. $$ \quad-4 $$ b. 3 or -4 c. $$ \frac{-1 \pm \sqrt{73}}{2} $$ d. 2 9. $$ \frac{2}{3 m}+\frac{4}{5 m}=2 $$ a. $$ \frac{22}{15} $$ b. $$ \frac{13}{15} $$ c. $$ \frac{11}{15} $$ d. $$ \frac{3}{8} $$ 10. $$ \frac{5}{x+3}=\frac{-2}{x+4} $$ a. $$ -\frac{26}{5} $$ b. $$ -\frac{6}{7} $$ c. $$ -\frac{10}{7} $$ d. $$ -\frac{26}{7} $$

Question

6. $$ \frac{x+2}{x-1} \div \frac{x+4}{x^{2}+4 x-5} $$ a. $$ \frac{(x+2)(x+5)}{x+4}, x 
eq-5,-4 $$ c. $$ \frac{(x+2)(x+4)}{(x-1)^{2}(x+5)}, x 
eq 1,-5,-4 $$ b. $$ \frac{(x+2)(x+4)}{(x-1)^{2}(x+5)}, x 
eq 1,-5 $$ d. $$ \frac{(x+2)(x+5)}{x+4}, x 
eq 1,-4,-5 $$ 7. $$ \frac{x^{2}-4}{x-8} \div(x-2) $$ a. $$ \frac{(x-2)(x+2)}{x-8} $$ c. $$ \frac{x-2}{x-8} $$ b. $$ \frac{x-8}{x+2} $$ d. $$ \frac{x+2}{x-8} $$ 8. $$ \frac{6}{x^{2}-9}-\frac{1}{x-3}=1 $$ a. $$ \quad-4 $$ b. 3 or -4 c. $$ \frac{-1 \pm \sqrt{73}}{2} $$ d. 2 9. $$ \frac{2}{3 m}+\frac{4}{5 m}=2 $$ a. $$ \frac{22}{15} $$ b. $$ \frac{13}{15} $$ c. $$ \frac{11}{15} $$ d. $$ \frac{3}{8} $$ 10. $$ \frac{5}{x+3}=\frac{-2}{x+4} $$ a. $$ -\frac{26}{5} $$ b. $$ -\frac{6}{7} $$ c. $$ -\frac{10}{7} $$ d. $$ -\frac{26}{7} $$

UpStudy AI · Accepted Answer

Let's solve each problem step by step.

### Problem 6: 
$$
\frac{x+2}{x-1} \div \frac{x+4}{x^{2}+4x-5}
$$

**Step 1: Rewrite the division as multiplication by the reciprocal.**
$$
\frac{x+2}{x-1} \cdot \frac{x^{2}+4x-5}{x+4}
$$

**Step 2: Factor the quadratic expression $$x^{2}+4x-5$$.**
To factor $$x^{2}+4x-5$$, we look for two numbers that multiply to $$-5$$ and add to $$4$$. These numbers are $$5$$ and $$-1$$.
$$
x^{2}+4x-5 = (x+5)(x-1)
$$

**Step 3: Substitute the factored form back into the expression.**
$$
\frac{x+2}{x-1} \cdot \frac{(x+5)(x-1)}{x+4}
$$

**Step 4: Cancel the common factor $$(x-1)$$.**
$$
\frac{(x+2)(x+5)}{x+4}
$$

**Step 5: Identify restrictions.**
The restrictions are $$x \neq -5$$ and $$x \neq -4$$ (from the denominator).

**Final Answer for Problem 6:**
$$
\frac{(x+2)(x+5)}{x+4}, \quad x \neq -5, -4
$$

### Problem 7:
$$
\frac{x^{2}-4}{x-8} \div (x-2)
$$

**Step 1: Rewrite the division as multiplication by the reciprocal.**
$$
\frac{x^{2}-4}{x-8} \cdot \frac{1}{x-2}
$$

**Step 2: Factor $$x^{2}-4$$ (difference of squares).**
$$
x^{2}-4 = (x-2)(x+2)
$$

**Step 3: Substitute the factored form back into the expression.**
$$
\frac{(x-2)(x+2)}{x-8} \cdot \frac{1}{x-2}
$$

**Step 4: Cancel the common factor $$(x-2)$$.**
$$
\frac{x+2}{x-8}
$$

**Final Answer for Problem 7:**
$$
\frac{x+2}{x-8}
$$

### Problem 8:
$$
\frac{6}{x^{2}-9}-\frac{1}{x-3}=1
$$

**Step 1: Factor $$x^{2}-9$$ (difference of squares).**
$$
x^{2}-9 = (x-3)(x+3)
$$

**Step 2: Rewrite the equation.**
$$
\frac{6}{(x-3)(x+3)} - \frac{1}{x-3} = 1
$$

**Step 3: Find a common denominator, which is $$(x-3)(x+3)$$.**
$$
\frac{6 - (x+3)}{(x-3)(x+3)} = 1
$$

**Step 4: Simplify the numerator.**
$$
\frac{6 - x - 3}{(x-3)(x+3)} = 1 \implies \frac{3 - x}{(x-3)(x+3)} = 1
$$

**Step 5: Cross-multiply.**
$$
3 - x = (x-3)(x+3)
$$

**Step 6: Expand the right side.**
$$
3 - x = x^2 - 9
$$

**Step 7: Rearrange the equation.**
$$
x^2 + x - 12 = 0
$$

**Step 8: Factor the quadratic.**
$$
(x+4)(x-3) = 0
$$

**Step 9: Solve for $$x$$.**
$$
x = -4 \quad \text{or} \quad x = 3
$$

**Final Answer for Problem 8:**
$$
3 \text{ or } -4
$$

### Problem 9:
$$
\frac{2}{3m}+\frac{4}{5m}=2
$$

**Step 1: Find a common denominator, which is $$15m$$.**
$$
\frac{10}{15m} + \frac{12}{15m} = 2
$$

**Step 2: Combine the fractions.**
$$
\frac{22}{15m} = 2
$$

**Step 3: Cross-multiply.**
$$
22 = 30m
$$

**Step 4: Solve for $$m$$.**
$$
m = \frac{22}{30} = \frac{11}{15}
$$

**Final Answer for Problem 9:**
$$
\frac{11}{15}
$$

### Problem 10:
$$
\frac{5}{x+3}=\frac{-2}{x+4}
$$

**Step 1: Cross-multiply.**
$$
5(x+4) = -2(x+3)
$$

**Step 2: Expand both sides.**
$$
5x + 20 = -2x - 6
$$

**Step 3: Rearrange the equation.**
$$
5x + 2x = -6 - 20
$$
$$
7x = -26
$$

**Step 4: Solve for $$x$$.**
$$
x = -\frac{26}{7}
$$

**Final Answer for Problem 10:**
$$
-\frac{26}{7}
$$

### Summary of Answers:
6. $$ \frac{(x+2)(x+5)}{x+4}, x \neq -5, -4 $$  
7. $$ \frac{x+2}{x-8} $$  
8. $$ 3 \text{ or } -4 $$  
9. $$ \frac{11}{15} $$  
10. $$ -\frac{26}{7} $$
6. $$ \frac{(x+2)(x+5)}{x+4}, x \neq -5, -4 $$ 7. $$ \frac{x+2}{x-8} $$ 8. $$ 3 \text{ or } -4 $$ 9. $$ \frac{11}{15} $$ 10. $$ -\frac{26}{7} $$

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c.
b.
d.

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c.
b.
d.

a.
b. 3 or -4
c.
d. 2

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b.
c.
d.

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b.
c.
d.

Upstudy AI Solution

Answer

Solution

Problem 6:

Problem 7:

Problem 8:

Problem 9:

Problem 10:

Summary of Answers:

Beyond the Answer

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