Problem 75 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of values of \(k\), for which the system of equations \((k+1) x+8 y=4 k\) \(k x+(k+3) y=3 k-1\) has no solution, is (A) 1 (B) 2 (C) 3 (D) infinite

Step-by-Step Solution

Answer

The number of values of \(k\) is 1.

Step 1: Understand the Condition for No Solution

A system of linear equations has no solution when the lines represented by the equations are parallel but not coincident. This occurs when the ratios of the coefficients of \(x\) and \(y\) in both equations are equal, but the ratio of the constant terms is different.

Step 2: Set the Equations into Standard Form

Rewrite the given system of equations in the standard form \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\). The equations are already in this format: 1. \((k+1)x + 8y = 4k\) 2. \(kx + (k+3)y = 3k - 1\)

Step 3: Determine the Ratios for Parallel Lines

For the lines to be parallel, their coefficients must satisfy: \[\frac{a_1}{a_2} = \frac{b_1}{b_2} eq \frac{c_1}{c_2}\] Substitute the respective coefficients from the equations: \[\frac{k+1}{k} = \frac{8}{k+3}\]

Step 4: Solve for the Condition of Ratios

Cross-multiply to solve for \(k\): \[(k+1)(k+3) = 8k\] Expand and simplify:\[k^2 + 3k + k + 3 = 8k\] \[k^2 + 4k + 3 = 8k\]Re-arrange to:\[k^2 - 4k + 3 = 0\] Factorize the quadratic equation:\[(k-3)(k-1) = 0\]Thus, \(k = 3\) or \(k = 1\).

Step 5: Verify the Constant Terms Condition

Substitute \(k = 3\) and \(k = 1\) into the constant term ratio:- For \(k = 3:\) \[\frac{4k}{3k-1} = \frac{12}{8} = \frac{3}{2}\] Check if not equal to \(\frac{8}{k+3} = \frac{8}{6} = \frac{4}{3}\).- For \(k = 1:\) \[\frac{4}{2} = 2\], compare with \(\frac{8}{4} = 2\). Not valid.Thus, only \(k = 3\) satisfies all conditions.

Key Concepts

Understanding No Solution Condition

In a system of linear equations, a "no solution" condition arises when the equations represent parallel lines that never intersect. These lines are close but never meet. It's like two paths running beside each other endlessly but never crossing. This is crucial in identifying when a system has no solutions.

To check for a no solution condition, compare the ratios of the coefficients and constants. For two linear equations, if the ratios of the coefficients of the variables are equal, while the ratio of the constant terms is different, then the system has no solution. In mathematical terms, for equations of the form \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\):

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

In our problem, finding the right \(k\) requires identifying these conditions to ensure the lines are parallel but non-coincident.

Recognizing Parallel Equations

Parallel equations arise when two lines move in the same direction at the same slope but differ in their positions on the graph. They stay equidistant from each other and never intersect. This property directly ties to the condition for no solution in a system.

In the standard form of linear equations, two lines are parallel when their coefficients form the same ratio:

• The coefficient of \(x\) in the first equation and the coefficient of \(x\) in the second equation are related by the same constant as the coefficients of \(y\).

For equations \((k+1)x + 8y = 4k\) and \(kx + (k+3)y = 3k-1\), this means matching:

• \(\frac{k+1}{k} = \frac{8}{k+3}\)

Solving this will determine whether the lines represented by these equations are indeed parallel, thus having no points of intersection.

Applying Quadratic Factorization

Quadratic factorization plays a significant role in simplifying equations and finding solutions for variables. It transforms polynomials into a product of simpler expressions, making it easier to solve quadratic equations for specific values of unknowns.

In our study of the condition for no solutions, we need to solve the quadratic equation derived from parallel conditions:

\[ k^2 - 4k + 3 = 0 \] Using factorization, break it down as:

• \((k-3)(k-1) = 0\)

This gives us possible values for \(k\):

• \(k = 3\)
• \(k = 1\)

Through factorization, we identify candidates for \(k\) that need to be verified against other conditions, such as the mismatched constant term ratios. This paves the way to understanding which values ensure that the system of equations has no solutions.