Problem 140 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

SYSTEM OF LINEAR EQUATIONS WITH PARAMETERS. $$ \left\\{\begin{array}{l} a x-y=b \\ b x+y=a \end{array}\right\\} $$

Step-by-Step Solution

Answer

The solution to the system of linear equations with parameters is \( x = 1, y = 0 \) for all \( a \neq b \) and \( a \neq -b \).

Step 1:Add the two equations

Add the equations \(a x - y = b\) and \(b x + y = a\) to remove the variable y. This gives the equation \( (a + b) x = (a + b) \). Divide both sides by \(a + b\) to isolate variable x, and we get the equation \( x = 1 \) provided that \(a + b \neq 0\).

Step 2: Subtract the two equations

Subtract the equations \(a x - y = b\) and \(b x + y = a\) to remove the variable y. This gives the equation \( (a - b) x = b - a \). Divide both sides by \(a - b\) to isolate variable x, which yields \( x = \frac{-1}{-1} = 1\) provided that \(a - b \neq 0\).

Step 3: Check for the consistency of solutions

If we compare the solutions from step 1 and 2, we see that, for the system of equations to have a solution, it is necessary that \(a + b \neq 0\) and \(a - b \neq 0\). In other words, the solutions are consistent unless \(a = b\) or \(a = -b\) meaning that the parameters a and b cannot be equal or opposite to each other. Hence, the solution for the system of equations is \( x = 1, y = 0 \) for all \( a \neq b \) and \( a \neq -b \).

Key Concepts

Linear Equations with Parameters

Linear equations with parameters introduce additional variables, often represented by letters, into the equation. These parameters can affect the overall solution of the system. Consider our example:

• Equation 1: \( ax - y = b \)
• Equation 2: \( bx + y = a \)

Here, the letters \( a \) and \( b \) are parameters. They are not fixed numbers but are treated like constants while solving. This means the same methods to solve equations with numbers apply, but with the complexity of considering multiple values for \( a \) and \( b \).
When solving these systems, you generally try to express one variable in terms of another or eliminate one of the variables, as shown in the process of equation addition and subtraction. Pay attention to conditions like \( a + b eq 0 \) because these ensure we don't divide by zero during our solving technique.

Solutions of Linear Systems

The solution to a system of linear equations is a set of values for the variables that fulfill all the equations simultaneously. In this case, solving meant finding \( x \) and \( y \) such that:

• From the first step, adding equations resulted in: \( (a + b)x = (a + b) \), simplifying to \( x = 1 \) if \( a + b eq 0 \).
• From the second step, subtracting equations gave \( (a - b)x = b - a \), which also simplifies to \( x = 1 \) if \( a - b eq 0 \).

Imagine this as tuning two knobs in a machine so both operate in harmony. If parameters \( a \) and \( b \) allow, you attain this consistent solution, ensuring each equation holds true. This set \( x = 1 \), when combined with another step, provides the comprehensive solution of \( x = 1 \) and \( y = 0 \).

Consistency of Solutions

A system of equations is considered consistent if there is at least one set of solutions that satisfies all equations. In our example, to ensure consistency:

• The sum \( a + b eq 0 \) is necessary to avoid zero in the denominator after adding the equations.
• The difference \( a - b eq 0 \) is essential for the same reason when equations are subtracted.

If these conditions aren't met, you might end up with contradictions or infinite solutions. For instance, if \( a = b \) or \( a = -b \), either equation loses defining power and can lead to:

• No solution, since the equations can contradict each other.
• Infinite solutions, if they symbolize the same line graphically.

However, for \( a eq b \) and \( a eq -b \), the system is consistent, leading to the exact solutions \( x = 1 \) and \( y = 0 \). This ensures parameters do not overlap in ways that would render the system unsolvable or overly complex.