Question

watch out for that intersection practice problems(online)
use the desmos graphing calculator to determine which pair of linear equations has the ordered pair (-6, 2) as its solution.
desmos | graphing calculator | texas version

Explanation:

Since the problem is about finding which pair of linear equations has the ordered pair \((-6, 2)\) as a solution using Desmos, but the specific pairs of linear equations are not provided here. However, the general method is as follows:

Step 1: Recall the definition of a solution to a system of linear equations

A solution \((x, y)\) to a system of linear equations satisfies both equations. So for the ordered pair \((-6, 2)\), we substitute \(x=-6\) and \(y = 2\) into each pair of linear equations (which are not given here, but would be in the original problem set) and check if both equations are true.

Step 2: Use Desmos (if equations are provided)

If we have the equations, we can graph them on Desmos. The intersection point of the two lines representing the linear equations will be the solution to the system. We check if the intersection point is \((-6, 2)\).

Since the specific pair of linear equations is not provided in the user's question (only the instruction to use Desmos to find which pair has \((-6,2)\) as a solution), we can't give the exact pair. But the process is to substitute \(x=-6\) and \(y = 2\) into each candidate pair of equations (or graph them and check the intersection) to see which pair is satisfied.

If we assume that we have a set of candidate pairs (for example, suppose we have pairs like \(y=x + 8\) and \(y=-\frac{1}{3}x\); let's test these):

For \(y=x + 8\), substitute \(x=-6\), \(y=2\): \(2=-6 + 8=2\) (true). For \(y=-\frac{1}{3}x\), substitute \(x=-6\), \(y = 2\): \(2=-\frac{1}{3}\times(-6)=2\) (true). So this pair \(

$$\begin{cases}y=x + 8\\y=-\frac{1}{3}x\end{cases}$$

\) would have \((-6,2)\) as a solution. But this is just an example.

Answer:

To determine the pair, substitute \(x = - 6\) and \(y=2\) into each pair of linear equations (or graph them on Desmos and check the intersection). The pair of equations where both equations are satisfied (or whose graphs intersect at \((-6,2)\)) is the solution. (Since the specific equations are not provided, the general method is as above. If equations were provided, we would test them and give the pair.)