Solving a Pair of Non-Linear Equations Through Linearization: A Comprehensive Guide

Non-linear equations can often be challenging to solve directly, but by using algebraic substitutions, we can transform them into a more manageable form—linear equations. This guide will walk you through the process of solving the given pair of non-linear equations through linearization. We will use a systematic approach to simplify the problem and find the solutions step-by-step.

Understanding the Problem

The given pair of non-linear equations is:

(frac{5}{x-1} frac{1}{y-2} - 2 0) (frac{6}{x-1} - frac{3}{y-2} - 1 0)

Step 1: Substitution

To simplify these equations, we introduce substitutions:

Let (u x - 1) Let (v y - 2)

With these substitutions, the equations become:

(frac{5}{u} frac{1}{v} - 2 0) (frac{6}{u} - frac{3}{v} - 1 0)

Step 2: Rewriting as Linear Equations

Next, we rearrange and rewrite these equations in terms of (u) and (v):

Rearranging the First Equation

(frac{5}{u} frac{1}{v} - 2 0) (frac{5}{u} frac{1}{v} 2) Multiply through by (uv): (5v u 2uv) Rearrange to standard form: (2uv - 5v - u 0)

Rearranging the Second Equation

(frac{6}{u} - frac{3}{v} - 1 0) (frac{6}{u} - frac{3}{v} 1) Multiply through by (uv): (6v - 3u uv) Rearrange to standard form: (uv - 6v - 3u 0)

Step 3: Forming the Linear System

Now, we have the following system of equations in (u) and (v):

(2uv - 5v - u 0) (uv - 6v - 3u 0)

Step 4: Solving the System

Let's express these equations in standard form:

(-u 2uv - 5v 0) (3u uv - 6v 0)

Factor out (u) and (v) from each equation:

From the first equation: (u(2v - 1) 5v) If (2v - 1 eq 0), then: (u frac{5v}{2v - 1})

Substitute (u frac{5v}{2v - 1}) into the second equation:

(3 left(frac{5v}{2v - 1}right) v left(frac{5v}{2v - 1}right) - 6v 0) Multiply through by (2v - 1): (15v 5v^2 - 12v^2 - 6v 0) Simplify: (-7v^2 9v 0) Factor out (v): (v(9 - 7v) 0) This gives us: (v 0) or (v 3)

Step 5: Finding Corresponding (u)

For each value of (v), we find the corresponding (u):

If (v 0): (u frac{5 cdot 0}{2 cdot 0 - 1} 0) If (v 3): (u frac{5 cdot 3}{2 cdot 3 - 1} frac{15}{6 - 1} 3)

Step 6: Finding (x) and (y)

Now we have two pairs ((u, v)):

((0, 0)) ((3, 3))

Converting these back to the original variables ((x, y)):

If (u 0): (x - 1 0), so (x 1); (v 0), so (y - 2 0), thus (y 2) If (u 3): (x - 1 3), so (x 4); (v 3), so (y - 2 3), thus (y 5)

Final Solutions

The solutions to the original equations are:

((x, y) (1, 2)) ((x, y) (4, 5))

Therefore, the solution set is:

boxed{(1, 2) and (4, 5)}

By transforming the non-linear equations into a system of linear equations, we can easily find the solutions. This method is particularly useful for handling complex algebraic systems.