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.

We will discuss coefficients in the general equation:

\(\displaystyle ax^2 +bx+c\)

In this case, \(\displaystyle c\) is positive and \(\displaystyle b\) is negative, and \(\displaystyle a=1\), so we know our answer involves two negative numbers that are factors of \(\displaystyle 6\) and add to \(\displaystyle -7\). The answer is:

\(\displaystyle (x-6)(x-1)\)

This is a factoring problem so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract \(\displaystyle -28\) from both sides to get \(\displaystyle x^{2}+16x+28=0\)

Think of the equation in this format to help with the following explanation.\(\displaystyle ax^{2}+bx+c=0\)

We must then factor to find the solutions for \(\displaystyle x\). To do this we must make a factor tree of \(\displaystyle c\) which is 28 in this case to find the possible solutions. The possible numbers are \(\displaystyle 1\cdot 28\), \(\displaystyle 2\cdot 14\), \(\displaystyle 4\cdot 7\).

Since \(\displaystyle c\) is positive we know that our factoring will produce two positive numbers.

We then use addition with the factoring tree to find the numbers that add together to equal \(\displaystyle b\). So \(\displaystyle 28+1=29\),  \(\displaystyle 2+14=16\), and \(\displaystyle 4+7=11\)

Success! 14 plus 2 equals \(\displaystyle b\). We then plug our numbers into the factored form of \(\displaystyle (x+2)(x+14)=0\)

We know that anything multiplied by 0 is equal to 0 so we plug in the numbers for \(\displaystyle x\) which make each equation equal to 0 so in this case \(\displaystyle x=-2,-14\).

By factoring one gets

\(\displaystyle y = (x-3)\left ( x + 5 \right )\)

Now setting each of the two factors to 0 (using the zero property), one gets

\(\displaystyle x= 3\) or \(\displaystyle x= -5\)