Each of the tables provided contains sets of ordered pairs. The input column represents the x-variables, and the output column represents the y-variables. We can tell if a set of ordered pairs represents a function when we match x-values to y-values. 

In order for a table to represents a function, there must be one and only one input for every output. This means that our correct answer will have all unique input values:

Functions cannot have more than one input value that is the same; thus, the following tables do not represent a function: 

In order to determine if an equation defines a linear function, we want to make sure that the equation of the line is in slope-intercept form:

\(\displaystyle y=mx+b\)

If we are unable to put an equation in this form, then the equation is not linear. 

Let's take a look at our answer choices:

\(\displaystyle 3y=9x^3-36\) 

Notice that in this equation our \(\displaystyle x\) value is to the third power, which does not match our slope-intercept form. 

\(\displaystyle x^2-8=y+21\)

Though this equation is not written in \(\displaystyle y=\) form, we can tell straight away that this does not define a linear function because the \(\displaystyle x\) value is to the second power. 

\(\displaystyle 5x^2=25y+10\)

Again, though this equation is not written in \(\displaystyle y=\) form, we can tell straight away that this does not define a linear function because the \(\displaystyle x\) value is to the second power. 

\(\displaystyle 3y+12x=21\)

For this equation, we can solve for \(\displaystyle y\) to make sure this equation can be written is slope-intercept form. From first glance it looks to be correct because none of our variables are written to a power. In order to tell for certain, we need to isolate the y variable on the left side of the equation.

First, we can subtract \(\displaystyle 12x\) from both sides:

\(\displaystyle \frac{\begin{array}[b]{r}3y+12x=21\\ -12x-12x\end{array}}{\\\\3y=-12x+21}\)

Next, we can divide each side by \(\displaystyle 3\textup:\)

\(\displaystyle \frac{3y}{3}=\frac{-12x+21}{3}=-4x+7\)

\(\displaystyle y=-4x+7\)

This equation is in slope-intercept form; thus, \(\displaystyle 3y+12x=21\) is the correct answer. 

There are a couple of ways to solve a system of linear equations: graphically and algebraically. In this lesson, we will review the two ways to solve a system of linear equations algebraically: substitution and elimination. 

Substitution can be used by solving one of the equations for either \(\displaystyle x\) or \(\displaystyle y\), and then substituting that expression in for the respective variable in the second equation. You could also solve both equations so that they are in the \(\displaystyle y=\) form, and then set both equations equal to each other. 

Elimination is best used when one of the variables has the same coefficient in both equations, because you can then use addition or subtraction to cancel one of the variables out, and solve for the other variable. 

For this problem, substitution makes the most sense because the first equation is already solved for a variable. We can substitute the expression that is equal to \(\displaystyle x\), into the \(\displaystyle x\) of our second equation:

\(\displaystyle 6y+3(1+3y)=-27\)

Next, we need to distribute and combine like terms:

\(\displaystyle 6y+3+9y=-27\)

\(\displaystyle 15y+3=-27\)

We are solving for the value of \(\displaystyle y\), which means we need to isolate the \(\displaystyle y\) to one side of the equation. We can subtract \(\displaystyle 3\) from both sides:

\(\displaystyle \frac{\begin{array}[b]{r}15y+3=-27\\ \ -3\ \ \ \ -3\end{array}}{\\\\15y=-30}\)

Then divide both sides by \(\displaystyle 15\) to solve for \(\displaystyle y\text:\)

\(\displaystyle \frac{15y}{15}=\frac{-30}{15}\)

\(\displaystyle y=-2\)

Remember, when we are solving a system of linear equations, we are looking for the point of intersection; thus, our answer should have both \(\displaystyle x\) and \(\displaystyle y\) values. 

Now that we have the value of \(\displaystyle y\), we can plug that value into the \(\displaystyle y\) variable in one of our given equations and solve for \(\displaystyle x\textup:\)

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

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

\(\displaystyle x=-5\)

Our point of intersection, and the solution to the system of linear equations is \(\displaystyle (-5,-2)\).