(3k+1)x+3y-2=0 (k 2+1)x+(k-2)y-5=0

4 min read Jun 16, 2024

Solving a System of Linear Equations with Parameters

This article will explore how to solve a system of linear equations involving parameters. We will focus on the following system:

(3k+1)x + 3y - 2 = 0

(k^2 + 1)x + (k-2)y - 5 = 0

We will use the method of elimination to solve for x and y in terms of k.

Understanding the Problem

The presence of the parameter 'k' makes this system slightly more complex than a typical system of linear equations. Our goal is to find the values of x and y that satisfy both equations for any given value of k.

Elimination Method

Multiply the equations to make the coefficients of one variable identical. We can achieve this by multiplying the first equation by (k-2) and the second equation by -3:

(3k+1)(k-2)x + 3(k-2)y - 2(k-2) = 0

-3(k^2 + 1)x - 3(k-2)y + 15 = 0
Add the equations together. This will eliminate the 'y' term:

(3k+1)(k-2)x - 3(k^2 + 1)x - 2(k-2) + 15 = 0
Simplify and solve for x. Expanding the terms and combining like terms:

(3k^2 - 5k - 2)x - 2k + 19 = 0

x = (2k - 19) / (3k^2 - 5k - 2)
Substitute the value of x back into either original equation to solve for y. Let's use the first equation:

(3k+1) * [(2k - 19) / (3k^2 - 5k - 2)] + 3y - 2 = 0
Solve for y. Simplifying and rearranging:

y = [(2 - (3k+1)(2k-19)) / (3k^2 - 5k - 2)] * (1/3)

Analyzing the Solution

We have found that:

x = (2k - 19) / (3k^2 - 5k - 2)
y = [(2 - (3k+1)(2k-19)) / (3k^2 - 5k - 2)] * (1/3)

These expressions give us the solution for x and y in terms of k.

Important Notes:

The denominator of both expressions is 3k^2 - 5k - 2. This denominator cannot be zero because division by zero is undefined. Therefore, we must exclude any value of k that makes the denominator equal to zero. We can factor the denominator to find these excluded values: (k - 2)(3k + 1) = 0. This means k cannot equal 2 or -1/3.
For all other values of k, the system has a unique solution.

Conclusion

We successfully solved the system of linear equations involving parameters using the elimination method. The solution for x and y is expressed in terms of k, providing a general solution that works for any value of k except for the excluded values that make the denominator zero. This approach allows us to understand the behavior of the system for different values of the parameter.