Analyzing Inconsistent Linear Equations

We are given a pair of linear equations and told they are inconsistent. We need to find a value that the sum of the parameters, \((k+p)\), cannot be.

Understanding Inconsistent Systems

A pair of linear equations in two variables, say \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), is considered inconsistent if it has no solution. This happens when the lines represented by these equations are parallel and distinct. The condition for a pair of linear equations to be inconsistent is:

\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Applying the Inconsistency Condition

The given equations are:

1. \(2x + ky = 5\)
2. \(4x + 6y = p\)

Comparing these with the standard form \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we have:

• \(a_1 = 2\), \(b_1 = k\), \(c_1 = 5\)
• \(a_2 = 4\), \(b_2 = 6\), \(c_2 = p\)

For the pair of linear equations to be inconsistent, the condition \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) must hold.

Let's substitute the values from our equations into the condition:

\(\frac{2}{4} = \frac{k}{6} \neq \frac{5}{p}\)

Solving for k and the Restriction on p

From the first part of the condition, \(\frac{2}{4} = \frac{k}{6}\):

\(\frac{1}{2} = \frac{k}{6}\)

Cross-multiplying gives:

\(1 \times 6 = 2 \times k\)

\(6 = 2k\)

\(k = \frac{6}{2}\)

\(k = 3\)

So, the value of \(k\) must be 3.

Now consider the second part of the condition, \(\frac{1}{2} \neq \frac{5}{p}\):

This inequality means the ratios \(\frac{a_1}{a_2}\) (or \(\frac{b_1}{b_2}\)) and \(\frac{c_1}{c_2}\) are not equal. Cross-multiplying the inequality gives:

\(1 \times p \neq 2 \times 5\)

\(p \neq 10\)

So, the value of \(p\) cannot be 10.

Determining the Value (k + p) Cannot Be

We found that \(k = 3\) and \(p \neq 10\).

The value of \((k + p)\) is \(3 + p\).

Since \(p\) cannot be 10, it follows that \(3 + p\) cannot be \(3 + 10\).

Therefore, \((k + p) \neq 13\).

Comparing with Options

The possible values for \((k + p)\) are given in the options: 11, 12, 13, and 15.

Our analysis shows that \((k + p)\) cannot be equal to 13.

Conclusion

For the given pair of linear equations to be inconsistent, \(k\) must be 3 and \(p\) must not be 10. This implies that the sum \((k + p)\) cannot be 13. We check the options and find that 13 is one of the given values.

Revision Table: Types of Pairs of Linear Equations

Additional Information: Solving Linear Equations

Pairs of linear equations can be solved using various methods:

• Graphical Method: Plotting both equations on a graph. The intersection point, if any, is the solution. Parallel lines mean no intersection (inconsistent), coincident lines mean infinite intersections (infinitely many solutions).
• Substitution Method: Solve one equation for one variable and substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method: Multiply one or both equations by suitable constants so that the coefficients of one variable become opposite or equal. Then add or subtract the equations to eliminate that variable.
• Cross-Multiplication Method: A specific method for solving systems of the form \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\).

Understanding the conditions for consistent and inconsistent systems helps determine the nature of solutions without actually solving the equations.