91Ó°ÊÓ

When we talk about systems of linear equations, we are referring to a set of two or more linear equations that we are trying to solve at the same time. The equations in a system are connected by their variables, and we're interested in finding the values for these variables that make all equations true simultaneously.

In the context of our exercise, we have a pair of equations which represent two straight lines. The point of intersection of these lines is the solution to the system since it's the point where both equations hold true. This solution can be one point, signify many points if the lines are the same, or there could be no intersection if the lines are parallel.

To visualize it, you can imagine each equation as a line on a graph. The intersection point is where the lines cross each other. However, we can also find this point algebraically without graphing, and that's what we're focusing on in this exercise.

One method to solve systems of equations algebraically is the substitution method. This technique involves expressing one variable in terms of the other from one equation and then substituting this expression into the second equation.

In the given exercise, since we already have both equations solved for the same variable, y, we can set these two expressions equal to each other, allowing us to apply the substitution method effectively. Let's walk through how it works:

First, write down the two equations so they are solving for the same variable, as we've done with \( y = 2x + 6 \) and \( y = -x - 6 \). Then, since both right-hand sides are equal to y, you can equate them with each other. This method helps us find the x-coordinate of the intersection point, which we will use to find the y-coordinate by plugging it back into one of the original equations.

After setting up our equations for substitution, we can move onto solving the resulting linear equation. This process involves combining like terms and isolating the variable we're solving for.

In our exercise, after substituting, we combine like terms by collecting all the x terms on one side of the equation and all the constants on the other. This gives us \( 3x = -12 \), which we simplify by dividing each side by 3, resulting in the solution \( x = -4 \).

With this value of x, we can then solve for y by substituting \( x \) back into either of the original equations (since they should both yield the same y value if our work is correct). This is the heart of solving linear equations: manipulating the equations step by step until we isolate the variable, providing us with the solution.