To find the excluded values, find the $\displaystyle x$-values that make the denominator equal zero.

1) set the denominator equal to zero.

$\displaystyle x^{3}+27=0$

2) solve for $\displaystyle x$.

$\displaystyle x^{3}=-27$

$\displaystyle x=-3$

To find the excluded values of a algebraic fraction you need to find when the denominator is zero. To find when the denominator is zero you need to factor it. This denominator factors into 

$\displaystyle (x-7)(x-4)$

so this is zero when x=4,7 so our answer is 

$\displaystyle x\neq4,7$

Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept.  Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,

$\displaystyle x^2-4=0$

The expression $\displaystyle x^2-4$ is a difference of squares that can be factored as 

$\displaystyle (x-2)(x+2)=0$

Solving this for $\displaystyle x$ gives either $\displaystyle 2$ or $\displaystyle -2$.  That means either of these values will make our numerator equal zero.  We might be tempted to conclude that both are valid answers.  However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself.  That means that any values that make our denominator zero must be rejected.  Therefore we must also look at the denominator.

$\displaystyle x^2+5x+6=0$ 

The left side factors as follows

$\displaystyle (x+3)(x+2)=0$

This means that if $\displaystyle x$ is $\displaystyle -3$ or $\displaystyle -2$, we end up dividing by zero.  That means that $\displaystyle -2$ cannot be a valid solution, leaving $\displaystyle 2$ as the only valid answer.  Therefore only #3 is correct. 

This problem involves a scalar multiplication with a matrix. Simply distribute the negative three and multiply this value with every number in the 2 by 3 matrix. The rows and columns will not change.

$\displaystyle -3\begin{bmatrix} 2& 3 & 4\\ 4&-5 & 10 \end{bmatrix}=\begin{bmatrix} -6&-9 &-12 \\ -12& 15&-30 \end{bmatrix}$

You can begin by treating this equation just like it was:

$\displaystyle 5x=14y$

That is, you can divide both sides by $\displaystyle 5$:

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} = \frac{14}{5}\begin{bmatrix} 20 \\ 12 \end{bmatrix}$

Now, for scalar multiplication of matrices, you merely need to multiply the scalar by each component:

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} \frac{14}{5}*20 \\ \frac{14}{5}*12 \end{bmatrix}$

Then, simplify:

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 56 \\ \frac{168}{5} \end{bmatrix}$

Therefore, $\displaystyle a+b=56+\frac{168}{5}=\frac{280}{5}+\frac{168}{5}=\frac{448}{5}$

Begin by distributing the fraction through the matrix on the left side of the equation. This will simplify the contents, given that they are factors of $\displaystyle 2$:

$\displaystyle \frac{1}{2}\begin{bmatrix} 2a\\4b \end{bmatrix}=\begin{bmatrix} a\\2b \end{bmatrix}$

Now, this means that your equation looks like:
$\displaystyle \begin{bmatrix} a\\2b \end{bmatrix}=\begin{bmatrix} 53 \\ 98 \end{bmatrix}$

This simply means:

$\displaystyle a=53$

and

$\displaystyle 2b=98$ or $\displaystyle b=49$

Therefore, $\displaystyle a+b=53+49=102$

Scalar multiplication and addition of matrices are both very easy. Just like regular scalar values, you do multiplication first:

$\displaystyle \begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + 4\begin{bmatrix} 5 & 3\\ 11 &22 \end{bmatrix}= \begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix}+\begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix}$

The addition of matrices is very easy. You merely need to add them directly together, correlating the spaces directly.

$\displaystyle \begin{bmatrix} 13 & 4\\ 71 &3 \end{bmatrix} + \begin{bmatrix} 20 & 12\\ 44 &88 \end{bmatrix} = \begin{bmatrix} 13 +20 & 4+12\\ 71+44 &3 +88\end{bmatrix}$

$\displaystyle \begin{bmatrix} 13 +20 & 4+12\\ 71+44 &3 +88\end{bmatrix}= \begin{bmatrix} 33 & 16\\ 115 & 91\end{bmatrix}$

When multplying any matrix by a scalar quantity (3 in our case), we simply multiply each term in the matrix by the scalar.

Therefore, every number simply gets multiplied by 3, giving us our answer.

$\displaystyle \begin{bmatrix} -9& 15&0 \\ 12& -3& 6 \end{bmatrix}$

The 3x3 identity matrix is 

$\displaystyle I = \begin{bmatrix} 1&0 &0 \\ 0&1&0 \\ 0&0 &1\end{bmatrix}$

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$\displaystyle n _{3,1} = m _{3,1} -6 i _{3,1}$

$\displaystyle m _{3,1}$ is the first element in the third row of $\displaystyle M$, which is 3; similarly, $\displaystyle i _{3,1} = 0$. Therefore, 

$\displaystyle n _{3,1} = 3 -6 (0)= 3$

The 3x3 identity matrix is 

$\displaystyle I = \begin{bmatrix} 1&0 &0 \\ 0&1&0 \\ 0&0 &1\end{bmatrix}$

Both scalar multplication of a matrix and matrix addition are performed elementwise, so

$\displaystyle n _{3,1} = 7 m _{3,1} + i _{3,1}$

$\displaystyle m _{3,1}$ is the first element in the third row of $\displaystyle M$, which is 3; similarly, $\displaystyle i _{3,1} = 0$. Therefore, 

$\displaystyle n _{3,1} = 7 (3) + 0 = 21$