Question

1. $\frac{x^{2}+2x+2}{x-1}=\frac{2x+3}{x-1}$
2. $\frac{2x}{x+2}=\frac{1}{x^{2}-4}+1$
3. $\frac{8}{2x+4}-\frac{3x+1}{x^{2}+2x}=\frac{2}{x+2}$
4. $\frac{x-24}{2x^{2}-9x-18}=\frac{x}{x-6}-\frac{2x}{2x+3}$

Explanation:

Problem 5

Step1: Eliminate common denominator

Since $x-1
eq 0$ (denominator can't be 0), multiply both sides by $x-1$:
$x^2 + 2x + 2 = 2x + 3$

Step2: Simplify the equation

Subtract $2x+3$ from both sides:
$x^2 + 2x + 2 - 2x - 3 = 0$
$x^2 - 1 = 0$

Step3: Solve for x

Factor and solve:
$(x-1)(x+1)=0$
$x=1$ or $x=-1$. Reject $x=1$ (makes denominator 0).

Step1: Factor denominators

$x^2-4=(x-2)(x+2)$, so $x
eq \pm2$. Multiply all terms by $(x-2)(x+2)$:
$2x(x-2) = 1 + (x-2)(x+2)$

Step2: Expand all terms

$2x^2 - 4x = 1 + x^2 - 4$

Step3: Simplify the equation

$2x^2 -4x -x^2 +3=0$
$x^2 -4x +3=0$

Step4: Solve quadratic equation

Factor:
$(x-1)(x-3)=0$
$x=1$ or $x=3$. Both are valid (no denominator zero).

Step1: Factor denominators

$2x+4=2(x+2)$, $x^2+2x=x(x+2)$, so $x
eq 0,-2$. Multiply all terms by $2x(x+2)$:
$8x - 2(3x+1) = 2 \cdot 2x$

Step2: Expand and simplify

$8x -6x -2 =4x$
$2x -2 =4x$

Step3: Solve for x

$-2x=2$
$x=-1$. Valid (no denominator zero).

Answer:

$x=-1$

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Problem 6