Problem 146 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the values of the parameters \(m\) and \(p\) such that the system \((3 m-5 p+1) x+(8 m-3 p-1) y=1\) \((2 m-3 p+1) x+(4 m-p) y=2\) possesses infinite solutions.

Step-by-Step Solution

Answer

The solution to this problem involves finding the values of \(m\) and \(p\) such that the given system of equations has infinite solutions. By setting up and solving the resulting system of equations, the values of \(m\) and \(p\) can be determined.

Step 1: Set up the Ratios

Set up the ratio of the coefficients of \(x\), \(y\), and the constant terms. This gives us: \(\frac{3m - 5p +1}{2m - 3p + 1} = \frac{8m - 3p - 1}{4m - p} = \frac{1}{2}\)

Step 2: Solve for the variables

We obtain two equations from the ratios as follows: \(\frac{3m - 5p +1}{2m - 3p + 1} = \frac{1}{2}\) and \(\frac{8m - 3p - 1}{4m - p} = \frac{1}{2}\). Solve these two equations to find the values of \(m\) and \(p\).

Key Concepts

Infinite Solutions

In a system of linear equations, infinite solutions occur when every equation in the system ultimately represents the same line. This means that there isn't just one intersection point or no intersection at all; instead, the lines lay on top of each other completely. When this happens, any point on this line is a solution to the system, hence the term "infinite solutions."

For a system to have infinite solutions, all the equations must be consistent and dependent. Consistency means there isn't any contradiction among them, and dependency indicates that these equations are basically multiples of one another. To achieve this, the ratios of corresponding coefficients of variables and constants must all be equal.

This is why we set up and solve for equal ratios of coefficients in our given problem. By ensuring this equality, we confirm that each equation stands for the same geometric line in a coordinate plane, providing infinite solutions.

Coefficient Ratios

The main technique for determining if a system of equations has infinite solutions involves comparing the ratios of coefficients, particularly those related to the variables and constant terms. The main idea here is to check whether these coefficients are proportionate. Let's break this down with a simple example:

Consider the two equations from our system:

• Equation 1: \( (3m-5p+1)x+(8m-3p-1)y=1\)
• Equation 2: \( (2m-3p+1)x+(4m-p)y=2 \)

To find if the lines are identical, you equate the coefficients of similar terms across both equations. That's where the concept of coefficient ratios comes in. In this problem, we equated the ratios of:

\(\frac{3m - 5p + 1}{2m - 3p + 1}\) for the x coefficients, \(\frac{8m - 3p - 1}{4m - p}\) for the y coefficients, and both equal to the ratio of the constant terms, \(\frac{1}{2}\).

This ensures that the equations represent the same line, leading to infinite solutions. The ratios tell us how equations are scaled against each other, and if the scaling is uniform throughout, the lines overlap.

Parameter Values

Finding the exact values for parameters means solving the equations resulting from the equal ratio conditions. Parameters, in this context like our variables \(m\) and \(p\), are elements within our equations whose values impact the nature of the solution.

Solving these gives us the specific values at which those equations are consistent and potentially dependent, resulting in infinite solutions.

The parameter values are determined based on the proportional relationships set up from the coefficient ratios. For instance, using given ratio equations:

• For \(\frac{3m - 5p + 1}{2m - 3p + 1} = \frac{1}{2}\): Solve it to find a relationship between \(m\) and \(p\)
• For \(\frac{8m - 3p - 1}{4m - p} = \frac{1}{2}\): Deduce another relationship

Solving these will yield values for \(m\) and \(p\) that satisfy both conditions, confirming the equations' fidelity when seen geometrically as one line. These values ensure the entire system behaves as required: infinite solutions through consistent and equivalent relations.