Find the values of 'a' and 'b' for which the system of equations 3x + y = 3 and (a - b)x + (a + b)y = 3a + b

Correct option is C

Given:

First equation: 3x + y = 3

Second equation: ((a - b)x + (a + b)y = 3a + b – 3

Concept Used:

To determine the nature of the solutions, we compare the ratios of the coefficients of the variables (x and y) and the constants.

For two equations:

a_1x + b_1y = c_1

a_2x + b_2y = c_2 ​

The conditions are:

Case	Condition (Ratios of coefficients)	Nature of Lines	Type of Solution
Unique solution	$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$	Intersecting lines	One unique solution
No solution	$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$	Parallel lines	No solution
Infinite solutions	$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$	Coincident lines	Infinitely many solutions

Solution:

Using the Ratio coefficients for infinite solution;

\frac{3}{a - b} = \frac{1}{a + b} = \frac{3}{3a + b - 3} ​

Solving,

\frac{3}{a - b} = \frac{1}{a + b}

3 (a + b) = a – b

3a + 3b = a – b

3a - a = -b - 3b

2a = -4b

a = -2b

Solving;

\frac{1}{a + b} = \frac{3}{3a + b - 3} ​

Substitute a = -2b into this equation:

\frac{1}{(-2b) + b} = \frac{3}{3(-2b) + b - 3}

\frac{1}{-b} = \frac{3}{-6b + b - 3} = \frac{3}{-5b - 3}​

(-5b - 3) = -3b

-5b + 3b = 3

b =

-\frac{3}{2}

Since a = -2b,

a =

-2 \times \left(-\frac{3}{2}\right) = 3

Thus, the values of a and b are:

\bf a = 3, \quad b = -\frac{3}{2}

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Assertion (A)	Reason (R)
The system of equations: $9x+6y=11$ and $7x+ky=9$ has no solution, if k = $\frac{14}{3}$ .	System of equations $ax+by=c$ $dx+ey=f$ has no solution, if $\frac{a}{d}=\frac{b}{e}\#\frac{c}{f}$

Find the values of 'a' and 'b' for which the system of equations 3x + y = 3 and (a - b)x + (a + b)y = 3a + b - 3 has infinite solutions.

a = 3, b = -2/3

a = 3, b = 2

a = 3, b = -3/2

a = 2, b = -3/2

Correct option is C

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