Factor out $\displaystyle -1$ from the numerator which gives us

$\displaystyle \left -(3-2x \right )$

Hence we get the following

$\displaystyle \frac{\left -(3-2x \right )}{3-2x}$

which is equal to $\displaystyle -1$

If we factors the denominator we get

$\displaystyle \left ( x+3 \right )^{2}$

Hence the rational expression becomes equal to

$\displaystyle \frac{\left ( x+3 \right )}{\left ( x+3 \right )^{2}}$

which is equal to $\displaystyle \frac{1}{\left ( x+3 \right )}$

We know that $\displaystyle -\frac{a}{b}$ is equivalent to $\displaystyle \frac{-a}{b}$ or $\displaystyle \frac{a}{-b}$.

By this property, there is no way to get $\displaystyle \frac{x+5}{2x+3}$ from $\displaystyle -\frac{x-5}{2x+3}$.

Therefore the correct answer is $\displaystyle \frac{x+5}{2x+3}$.

If $\displaystyle \left ( x-1 \right ) = 0$ or $\displaystyle \left ( x+2 \right ) =0$, the denominator is 0, which makes the expression undefined.

 This happens when x = 1 or when x = -2.

Therefore the correct answer is $\displaystyle x = 1, -2$.

The two fractions on the left side of the equation need a common denominator. We can easily do find one by multiplying both the top and bottom of each fraction by the denominator of the other.

 $\displaystyle \frac{y+1}{8y+2}\cdot \frac{y}{y}$  becomes $\displaystyle \frac{y^{2}+y}{(y)(8y+2)}$.

$\displaystyle \frac{3}{y}\cdot \frac{8y+2}{8y+2}$  becomes $\displaystyle \frac{24y+6}{(y)(8y+2)}$.

Now add the two fractions: $\displaystyle \frac{y^{2}+25y+6}{(y)(8y+2)}$

To solve, multiply both sides of the equation by $\displaystyle (y)(8y+2)$, yielding

 $\displaystyle y^{2}+25y+6 = \frac{(40y^{2}+10y)}{3}$.

Multiply both sides by 3:

 $\displaystyle 3y^{2}+75y+18 = 40y^{2}+ 10y$

Move all terms to the same side:

 $\displaystyle 37y^{2}-65y-18 = 0$

This looks like a complicated equation to factor, but luckily, the only factors of 37 are 37 and 1, so we are left with

 $\displaystyle (37y+9)(y-2)$.

Our solutions are therefore

$\displaystyle y=2$ 

and

$\displaystyle y=-\frac{9}{37}$.

First, factor out x from the numerator:

$\displaystyle \frac{x^{3}-5x^{2}+6x}{x-3}$$\displaystyle =\frac{x(x^{2}-5x+6)}{x-3}$

Notice that the resultant expression in the parentheses is quadratic. This expression can be further factored:

$\displaystyle =\frac{x(x-2)(x-3)}{x-3}$

We can then cancel the (x-3) which appears in both the numerator and denominator:

$\displaystyle =\frac{x(x-2)}{1}$

Finally, distribute the x outside of the parentheses to reach our answer:

$\displaystyle =x^{2}-2x$

Multiply each side by $\displaystyle 3y+2x^{2}$

$\displaystyle 2(3y+2x^{2})=10$

Distribute 2 to each term of the polynomial.

$\displaystyle 6y+4x^{2}=10$

Divide the polynomial by 6.

$\displaystyle 6(y+\frac{2}{3}x^{2})=10$

Divide each side by 6.

$\displaystyle y+\frac{2}{3}x^{2}=\frac{5}{3}$

Subtract the $\displaystyle x$ term from each side.

$\displaystyle y=\frac{5}{3}-\frac{2}{3}x^{2}$

Multiply each side by $\displaystyle (k+2)$

$\displaystyle 3(k+2)=10$

Distribute 3 to the terms in parentheses.

$\displaystyle 3k+6=10$

Subtract 6 from each side of the equation.

$\displaystyle 3k=4$

Divide each side by 3.

$\displaystyle k=\frac{4}{3}$

$\displaystyle 5=\frac{105}{n^{2}+5}$

Multiply each side of the equation by $\displaystyle n^{2}+5$

$\displaystyle 5(n^{2}+5)=105$

Distribute 5 to each term in parentheses.

$\displaystyle 5n^{2}+25=105$

Subtract 25 from each side of equation.

$\displaystyle 5n^{2}=80$

Divide each side of equation by 5.

$\displaystyle n^{2}=16$

Square root of each side of equation.

$\displaystyle n=4$

First, factor the numerator. We need factors that multiply to $\displaystyle \small -6$ and add to $\displaystyle \small -5$.

$\displaystyle \small 1*-6=-6\ \text{and}\ 1+(-6)=-5$

$\displaystyle \small x^2-5x-6=(x-6)(x+1)$

We can plug the factored terms into the original expression.

$\displaystyle \small \frac{x^2-5x-6}{x+1}=\frac{(x-6)(x+1)}{(x+1)}$

Note that $\displaystyle \small (x+1)$ appears in both the numerator and the denominator. This allows us to cancel the terms.

$\displaystyle \frac{(x-6)(x+1)}{(x+1)}=(x-6)$

This is our final answer.