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The substitution method is a tool in algebra used to find solutions to systems of equations. A system of equations consists of two or more equations with the same set of variables. The goal is to find the value for each variable that will make all equations true.

When using the substitution method, you first solve one of the equations for one variable in terms of the others. Next, you substitute this expression into the other equation(s), replacing the variable. This transforms the system of equations with several variables into one with fewer variables. Continue this process until you have a single equation with one variable, which you can then solve.

For example, in our exercise, we have two equations with the variable 'y' expressed in terms of 'x'. By substituting one equation into the other, we are able to find the value of 'x', transforming the system into a single-variable equation.

Linear equations are the building blocks of algebra. They are equations that involve variables raised to the first power and which graph as straight lines on a coordinate plane. A linear equation in two variables, such as 'x' and 'y', can be written in the form of \(y = mx + b\), where 'm' is the slope of the line and 'b' is the y-intercept.

In our exercise, both given equations are linear since they can be rearranged into this form. The solutions to these equations are the coordinates \(x, y\) that satisfy both equations simultaneously. In the case of the provided equations, \(y = -4x + 2\) and \(y = -3x - 1\), they represent two different lines, and their intersection is the solution to the system.

Finding algebraic solutions means solving equations or systems of equations to find the values of unknown variables that make the equations true. This involves a series of algebraic manipulations which include operations like addition, subtraction, multiplication, division, and substitution.

In the context of our substitution exercise, the algebraic solution involves manipulating the equations to isolate 'x' and then using this value to find 'y'. The steps include rearranging the terms, combining like terms, and performing arithmetic calculations in a logical order. The final result \(x = -3\) and \(y = 10\) from our exercise is the algebraic solution, providing the specific pair of values that satisfy both original equations in the system.