Question

1. solve: $\frac{2}{x-1}=\frac{4}{x+3}$
2. solve: $\frac{6}{x+5}=3$
3. solve: $\frac{1}{x}+\frac{3}{x^2}=\frac{4}{2x^2}$
4. what is the common denominator for: $\frac{x}{x^2 - 2x - 15}$ and $\frac{1}{x^2 - 25}$
5. find all extraneous solutions: $\frac{4}{x-1}+\frac{5}{x+3}=\frac{-1}{x}$
6. which equation would have an x-intercept of (7,0)?

a. $\frac{3x+1}{4}$
b. $\frac{4x-28}{x-2}$
c. $\frac{3x+24}{x-1}$
d. $\frac{x+1}{9}$

Explanation:

Problem 9

Step1: Cross-multiply to eliminate fractions

$2(x+3) = 4(x-1)$

Step2: Expand both sides

$2x + 6 = 4x - 4$

Step3: Rearrange to isolate $x$

$6 + 4 = 4x - 2x$
$10 = 2x$

Step4: Solve for $x$

$x = \frac{10}{2} = 5$

Problem 10

Step1: Multiply both sides by $x+5$

$6 = 3(x+5)$

Step2: Expand the right side

$6 = 3x + 15$

Step3: Rearrange to isolate $x$

$6 - 15 = 3x$
$-9 = 3x$

Step4: Solve for $x$

$x = \frac{-9}{3} = -3$

Problem 11

Step1: Multiply all terms by $2x^2$ (LCD)

$2x^2 \cdot \frac{1}{x} + 2x^2 \cdot \frac{3}{x^2} = 2x^2 \cdot \frac{4}{2x^2}$
$2x + 6 = 4$

Step2: Rearrange to isolate $x$

$2x = 4 - 6$
$2x = -2$

Step3: Solve for $x$

$x = \frac{-2}{2} = -1$

Problem 12

Step1: Factor denominators

$x^2-2x-15=(x-5)(x+3)$
$x^2-25=(x-5)(x+5)$

Step2: Identify common denominator

Combine unique factors: $(x-5)(x+3)(x+5)$

Problem 13

Step1: Multiply by LCD $x(x-1)(x+3)$

$4x(x+3) + 5x(x-1) = -1(x-1)(x+3)$

Step2: Expand all terms

$4x^2+12x + 5x^2-5x = -(x^2+2x-3)$
$9x^2+7x = -x^2-2x+3$

Step3: Rearrange to standard quadratic

$9x^2+7x+x^2+2x-3=0$
$10x^2+9x-3=0$

Step4: Check domain restrictions

Original equation has restrictions: $x
eq0,1,-3$. The quadratic solutions are $x=\frac{-9\pm\sqrt{81+120}}{20}=\frac{-9\pm\sqrt{201}}{20}$, none match restricted values.

Step5: Conclusion on extraneous solutions

No extraneous solutions.

Problem 14

Step1: Test $x=7$ for each option

For option a: $\frac{3(7)+1}{4}=\frac{22}{4}
eq0$
For option b: $\frac{4(7)-28}{7-2}=\frac{28-28}{5}=0$
For option c: $\frac{3(7)+24}{7-1}=\frac{45}{6}
eq0$
For option d: $\frac{7+1}{9}=\frac{8}{9}
eq0$

Answer:

1. $x=5$
2. $x=-3$
3. $x=-1$
4. $(x-5)(x+3)(x+5)$ (or $x^3+3x^2-25x-75$)
5. No extraneous solutions
6. b. $\frac{4x-28}{x-2}$