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The substitution method is a technique used to solve systems of linear equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation.
First, choose one of the equations in the system. It's often easiest to pick the simpler one. In this example, we chose the second equation: \(3x + y = 6\).
Next, solve this equation for one variable. Here, we solved for \(y\), giving us \(y = 6 - 3x\).
Then, substitute this expression into the other equation. By replacing \(y\) in the first equation \(8x - 10y = -22\) with \(6 - 3x\), we get a single equation in terms of \(x\) only. This substitution step simplifies the system and allows us to solve for one variable at a time.
Finally, once we've found the value for one variable, we can substitute it back into the expression we found earlier to find the value of the other variable.

Linear equations represent straight lines when graphed on a coordinate plane. They are typically in the form \(ax + by = c\).
In the given example, we have two linear equations: \(8x - 10y = -22\) and \(3x + y = 6\). These equations can be solved simultaneously because they intersect at a point. By finding this intersection point, we can determine the values of \(x\) and \(y\) that satisfy both equations.
Linear equations are fundamental in algebra. They are used to model relationships between variables. Systems of linear equations can have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (coinciding lines).
In our case, the system has a unique solution where the two lines intersect. This point of intersection gives the values of the variables that solves both equations simultaneously.

Solving for variables in a system of linear equations requires isolating each variable step by step. Let's break this down:

• Start with one equation and solve it for one of the variables. For instance, in the equation \(3x + y = 6\), we solved for \(y\), obtaining \(y = 6 - 3x\).
• Substitute this result into the other equation to eliminate \(y\). This gives us a single equation with one variable. In this case, substituting into \(8x - 10y = -22\) leads to \(8x - 10(6 - 3x) = -22\).
• Simplify and solve for the remaining variable. Here, we simplified to get \(38x - 60 = -22\) and then solved for \(x\). This gave us \(x = 1\).
• Once \(x\) is found, substitute it back into the equation from which you originally solved for \(y\). Substituting \(x = 1\) back into \(y = 6 - 3(1)\) results in \(y = 3\).

The solution is the set of values for the variables that makes both equations true. Hence, for this system, \(x = 1\) and \(y = 3\) is the solution.