Equation 1: \(\displaystyle x+y=4\)

Equation 2: \(\displaystyle x-y=8\)

Solve equation 2 for X: \(\displaystyle x=8+y\)

Substitute into Equation 1: \(\displaystyle (8+y)+y=4\)

Solve for y: \(\displaystyle 2y=-4\),  \(\displaystyle y=-2\)

Take the answer for y and plug it back into either original equation to find x:

\(\displaystyle x=6, y=-2\)

Equation 1: \(\displaystyle 2x+y=16\)

Equation 2: \(\displaystyle y-2x=0\)

Solve Equation 2 for y: 

\(\displaystyle y=2x\)

Substitute into Equation 1:

\(\displaystyle 2x+2x=16\)

\(\displaystyle 4x=16\)

\(\displaystyle x=4\)

Plug x back into either original equations and solve for y:

\(\displaystyle x=4,y=8\)

Simply plugging in -2 into each answer choice will determine the correct answer:

\(\displaystyle 6x^3 = 6(-2)^3 = -48\)

\(\displaystyle 2x-5=2(-2)-5=-9\)

\(\displaystyle 2x^2-13=2(-2)^2-13=-5\)

\(\displaystyle 5x^2 + 2 = 5(-2)^2 + 5 = 25\)

\(\displaystyle 2x-10 = 2(-2) - 10 = -14\)

The key to solving this problem is to remember the order of operations and that negative numbers squared are positive, while negative numbers raised to the third power are negative.

When solving the equation \(\displaystyle x^3 - 6 = 210\), observe that \(\displaystyle x^3 = 216\). Taking the cube root of 216 gives 6. Thus,  \(\displaystyle 2x + 3=2(6) + 3=15.\)

In the box, there are \(\displaystyle 7x\) kilograms of baseballs and \(\displaystyle 9x\) kilograms of footballs. In total, there are \(\displaystyle 16x\) kilograms of balls. The total weight of the box is 48 kilograms, so \(\displaystyle 16x = 48\: and\: x = 3.\) Since there are \(\displaystyle 9x\) kilograms of footballs, the total weight of the footballs in the box is equal to: \(\displaystyle 9(3) = 27.\)

To begin solving, we use the distributive property to simplify the left side. We get: \(\displaystyle 5x - 25 = 5\). Then, adding 25 to both sides, we get: \(\displaystyle 5x = 30\). Dividing by 5, we get that \(\displaystyle x = 6\). 

\(\displaystyle 4x + 2 = 3x - 1\)

Since we have variables on both sides, we first subtract \(\displaystyle 3x\) from both sides.

This leaves us with \(\displaystyle x + 2 = -1\).

Subtracting 2 from both sides, we get \(\displaystyle x = -3\). 

\(\displaystyle 4(3 - y) = 6y + 2\)

First, use the distributive property to multiply out the left side:

\(\displaystyle 12 - 4y = 6y + 2\)

Then, add \(\displaystyle 4y\) on both sides: 

\(\displaystyle 12 = 10y + 2\)

Subtract 2 from both sides: 

\(\displaystyle 10 = 10y\)

Divide both sides by 10 to get \(\displaystyle 1 = y\), or \(\displaystyle y = 1\). 

The first step in solving this equation is to distribute the 2 through the parentheses. This gives us:

\(\displaystyle 2\cdot z + 2\cdot 3 = 18\)

\(\displaystyle 2z+6=18\)

Next, we subtract 6 from both sides, in order to get the variable alone on one side of the equation:

\(\displaystyle 2z+6 - 6 =18-6\)

\(\displaystyle 2z = 12\)

Finally, we divide both sides by 2 to solve for \(\displaystyle z\):

\(\displaystyle \frac{2z}{2} = \frac{12}{2}\)

\(\displaystyle z = 6\)

The first step in solving this equation is to combine the like terms. That is, move all of the terms with an \(\displaystyle x\) in it to one side of the equation, and all the terms without an \(\displaystyle x\) to the other side. Let's begin with moving the \(\displaystyle x\)terms to the left side of the equation. We do this by subtracting \(\displaystyle 4x\) from each side:

\(\displaystyle 6x + 3 - 4x = 4x - 7 - 4x\)

\(\displaystyle 2x+3=-7\)

Next, we combine the terms without an \(\displaystyle x\) on the right side. We do this by subtracting 3 from both sides:

\(\displaystyle 2x+3-3 = -7-3\)

\(\displaystyle 2x = -10\)

Finally, we divide both sides of the equation by 2 to solve for \(\displaystyle x\):

\(\displaystyle \frac{2x}{2} = \frac{-10}{2}\)

\(\displaystyle x = -5\)