This problem is asking for the factors of 24 that are NOT also factors of 16 OR 30. Thus, the quickest way to solve this problem is to just list all the factors of 16, 24, and 30, and solve.  Those factors are:

16: 1, 2, 4, 8, 16

24: 1, 2, 3, 4, 6, 8, 12, 24

30: 1, 2, 3, 5, 6, 10, 15, 30

As you can see you can knock out 1 (16 & 30), 2 (16 & 30), 3 (30),  4 (16), 6 (30) and 8 (16) as factors of 24 that are in common with 16 & 30. This leaves only 12 and 24, which is the solution.

To find the solution set, you must solve the equation; in this case, solving the equation means isolating $\displaystyle k$ on one side of the equation, and the numbers on the other side of the equation.  

That is done like this:

$\displaystyle k^{2} - 24 = 56$

$\displaystyle k^{2} = 81$

$\displaystyle k = \pm 9$

K = -9 or 9 because either number is the square root of 81. To see that that's true, square both numbers.  $\displaystyle -9 \times-9 = 81$ and $\displaystyle 9 \times9=81$. 

This is very important to remember: whenever you're isolating a variable by taking the square root of a squared number, the answer can be a positive OR negative value, as long as they share an absolute value!

The perfect square trinomial in the equation can be factored to yield an equivalent equation as follows:

$\displaystyle x^{4} +2,000x^{2}+ 1,000,000 = 0$

$\displaystyle \left (x^{2} \right )^{2}+2 \cdot x^{2}\cdot 1,000+ 1, 000^{2} = 0$

$\displaystyle \left (x^{2} +1,000 \right ) ^{2}= 0$

$\displaystyle x^{2} +1,000 = 0$

$\displaystyle x^{2} = -1,000$

$\displaystyle x = \pm \sqrt{-1,000}$

Therefore, there are exactly two solutions to the equation, both imaginary.

First, since the equation is quadratic, put it in standard form 

$\displaystyle ax^{2} + bx + c = 0$

as follows:

$\displaystyle 4x^{2} + 7x - 5 = x^{2} + 2x - 10$

$\displaystyle 4x^{2} + 7x - 5 - x^{2} - 2x + 10 = x^{2} + 2x - 10 - x^{2} - 2x + 10$

$\displaystyle 3x^{2} +5x +5 = 0$

To determine the nature of the solution set, evaluate discriminant $\displaystyle b^{2}- 4ac$ for $\displaystyle a = 3, b= 5, c=5$:

$\displaystyle b^{2}- 4ac$

$\displaystyle = 5^{2}- 4 (3)(5)$

$\displaystyle = 25- 60$

$\displaystyle = -35$

The discriminant is negative, so the solution set comprises two imaginary numbers.

The most simple method for solving systems of equations is to transform one of the equations so it allows for the canceling out of a variable. In this case, we can multiply $\displaystyle 3x + y = 8$ by $\displaystyle (-4)$ to get $\displaystyle -12x - 4y = -32$.

 Then, we can add $\displaystyle 2x + 4y = 12$ to this equation to yield $\displaystyle -10x = -20$, so $\displaystyle x = 2$.

We can plug that value into either of the original equations; for example, $\displaystyle 3(2) )+ y = 8$.

So, $\displaystyle y = 2$ as well.

By solving one equation for $\displaystyle y$, and replacing $\displaystyle y$ in the other equation with that expression, you generate an equation of only 1 variable which can be readily solved.

The number of pounds of coffee beans totals 50, so one of the equations would be

$\displaystyle x + y = 50$.

The total price of the Kona beans, is its unit price, $24 per pound, multiplied by its quantity, $\displaystyle x$ pounds. This is $\displaystyle 24x$ dollars. Similarly, the total price of the Ethiopian delight beans is $\displaystyle 10y$ dollars, and the price of the mixture is $\displaystyle 14.20 \times 50 = 710$ dollars. Add the prices of the Kona and Ethiopian Delight beans to get the price of the mixture:

$\displaystyle 24x + 10 y = 710$

We are trying to solve for $\displaystyle x$ in the system

$\displaystyle 24x + 10 y = 710$

$\displaystyle x + y = 50$

Multiply the second equation by $\displaystyle -10$, then add to the first:

    $\displaystyle 24x + 10 y = 710$

$\displaystyle \underline{-10x -10 y = -500}$

$\displaystyle 14x$                  $\displaystyle = 210$

$\displaystyle x = 210 \div 14$

$\displaystyle x = 15$

The mixture includes 15 pounds of Kona beans.

To solve this problem, you must first solve the system of equations for both $\displaystyle x$ and $\displaystyle y$, then plug the values of $\displaystyle x$ and $\displaystyle y$ into the final equation.  

In order to solve a system of equations, you must add the equations in a way that gets rid of one of the variables so you can solve for one variable, then for the other. One example of how to do so is as follows:

Take the equations. Multiply the first equation by two so that there is $\displaystyle -2y$ (this will cancel out the $\displaystyle 2y$ in the second equation).

   $\displaystyle (3x-y=1) \times2 = 6x-2y=2$

Add the equations:

$\displaystyle 6x-2y=2$

$\displaystyle 2x+2y=14$

Find the sum (notice that the variable $\displaystyle y$ has disappeared entirely):

$\displaystyle 8x=16$

Solve for $\displaystyle x$.

$\displaystyle x = 2$

Plug this value of $\displaystyle x$ back into one of the original equations to solve for $\displaystyle y$:

$\displaystyle 3(2)-y=1$

$\displaystyle 6-y=1$

$\displaystyle y=5$

Now, plug the values of $\displaystyle x$ and $\displaystyle y$ into the final expression:

$\displaystyle 2+5 = 7$

The answer is $\displaystyle 7$. 

$\displaystyle 4x+3y=6$

$\displaystyle 2x+2y=4$

For the second equation, solve for $\displaystyle x$ in terms of $\displaystyle y$.

$\displaystyle 4-2x=2y$

$\displaystyle y=2-x$

Plug this value of y into the first equation.

$\displaystyle 4x+3(2-x)=6$

$\displaystyle 4x + 6- 3x =6$

$\displaystyle x=0$

In the second equation, you can substitute $\displaystyle 3x + 4$ for $\displaystyle y$ from the first.

$\displaystyle 2x + 3y = 34$

$\displaystyle 2x + 3 (3x + 4) = 34$

$\displaystyle 2x + 3 (3x) + 3 (4) = 34$

$\displaystyle 2x + 9x + 12 = 34$

$\displaystyle 11x + 12 = 34$

$\displaystyle 11x = 22$

$\displaystyle x = 2$

Now, substitute 2 for $\displaystyle x$ in the first equation:

$\displaystyle y = 3x + 4$

$\displaystyle y = 3 (2) + 4$

$\displaystyle y = 6 + 4$ 

$\displaystyle y = 10$

The solution is $\displaystyle (2, 10)$