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The substitution method is a powerful technique for solving systems of equations. It involves replacing one variable with an expression involving another variable. In this exercise, we use the substitutions \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). This helps to transform the original non-linear system into a simpler, linear system. By substituting these values into the original equations, we reshape the problem into a more solvable form.
For example, the initial equation \( -\frac{3}{x} + \frac{4}{y} = 11 \) transforms to \( -3u + 4v = 11 \) using our substitutions. This method simplifies the complexity, making it easier to solve for the variables in a sequential manner.

Algebraic manipulation involves using algebraic rules to rearrange and simplify expressions. Once we've substituted \( u \) and \( v \) into the system, we apply algebraic manipulation to isolate one of the variables.
In the second step, after substitution, we get the new equation \( u - 2v = -5 \). Solving for \( u \) yields \( u = 2v - 5 \).
By substituting \( u \) back into the first transformed equation, we perform algebraic operations like distribution and combining like terms. This transforms the equation into a simpler one that can be solved more easily. For instance, substituting \( u = 2v - 5 \) into \( -3u + 4v = 11 \) and simplifying gives us \( -6v + 15 + 4v = 11 \), and further manipulation leads us to find \( v = 2 \).

Back substitution is used after solving one of the simpler equations for one variable. The obtained value is substituted back into the equation to find the second variable.
For example, after finding \( v = 2 \), we substitute \( v = 2 \) back into the equation \( u = 2v - 5 \). This gives us \( u = 2(2) - 5 = 4 - 5 = -1 \). Back substitution allows us to determine the values of all variables sequentially.
This technique ensures that we systematically find all the unknowns in the system by working backward from the simpler, solved forms of the equations.

Linear equations are algebraic equations in which each term is either a constant or the product of a constant and a single variable. The system we derived from substitution \( -3u + 4v = 11 \) and \( u - 2v = -5 \) are linear equations. These types of equations are easier to solve because they don't involve variables raised to any powers other than one.
Once we convert the original non-linear equations to linear forms, using methods like solving for one variable, substituting into another, and combining like terms, we can efficiently reach the solution. Linear equations are foundational concepts in algebra and provide a stepping-stone for understanding more complex algebraic structures.