Finding the Value of k for Linear Equations with No Solution

The problem asks us to find the value of 'k' for which the given system of linear equations has no solution. We are given two linear equations in two variables, x and y:

1. \(6x - 5y + 11 = 0\)
2. \(15x + ky - 9 = 0\)

Understanding the Condition for No Solution

A system of two linear equations of the form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) is said to have no solution if the lines represented by these equations are parallel and distinct. This condition is met when the ratio of the coefficients of x is equal to the ratio of the coefficients of y, but this ratio is not equal to the ratio of the constant terms. Mathematically, this is expressed as:

Identifying Coefficients

From the given equations, we can identify the coefficients:

For equation 1, \(6x - 5y + 11 = 0\):

• \(a_1 = 6\) (coefficient of x)
• \(b_1 = -5\) (coefficient of y)
• \(c_1 = 11\) (constant term)

For equation 2, \(15x + ky - 9 = 0\):

• \(a_2 = 15\) (coefficient of x)
• \(b_2 = k\) (coefficient of y)
• \(c_2 = -9\) (constant term)

Applying the No Solution Condition

Using the condition for no solution, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), we substitute the coefficients:

To find the value of k, we use the equality part of the condition:

Solving for k

First, simplify the fraction \(\frac{6}{15}\). Both 6 and 15 are divisible by 3:

So, the equation becomes:

Now, we can solve for k by cross-multiplication:

Divide both sides by 2 to find k:

Verifying the Inequality Condition

We must also check if the inequality \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) holds for \(k = -12.5\). This means checking if \(\frac{-5}{-12.5} \neq \frac{11}{-9}\).

The first ratio is \(\frac{-5}{-12.5} = \frac{5}{12.5}\). To remove the decimal, multiply numerator and denominator by 10:

Both 50 and 125 are divisible by 25:

The second ratio is \(\frac{11}{-9} = -\frac{11}{9}\).

We need to check if \(\frac{2}{5} \neq -\frac{11}{9}\). Since \(\frac{2}{5}\) is a positive value and \(-\frac{11}{9}\) is a negative value, they are indeed not equal. Thus, the condition \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) is satisfied for \(k = -12.5\).

Conclusion

Therefore, the value of k for which the system of equations \(6x - 5y + 11 = 0\) and \(15x + ky - 9 = 0\) has no solution is -12.5.

Revision Table: System of Linear Equations Conditions

Additional Information on Inconsistent Systems

A system of linear equations that has no solution is called an inconsistent system. This means there is no pair of values (x, y) that can satisfy both equations simultaneously. When graphed, these equations represent lines that are parallel but do not overlap at any point. The distance between the two lines is constant, and they never intersect. The condition \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) ensures the lines are parallel (same slope), and the condition \(\neq \frac{c_1}{c_2}\) ensures they are distinct (different y-intercepts if \(b_1, b_2 \neq 0\)). Understanding these conditions is crucial for analyzing the nature of solutions for any system of two linear equations in two variables.