# For what values of a and b, the following system of equations have infinite number of solutions ?

## Question :

(i)  For what values of a and b, the following system of equations have infinite number of solutions ?

2x + 3y = 7

(a – b)x + (a + b)y = 3a + b – 2

(ii)  For what value of k will the following pair of linear equations have no solution ?

3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

## Solution :

(i)  The given linear equations can be written as

2x + 3y – 7 = 0

(a – b)x + (a + b)y – (3a + b – 2) = 0

Above equations is of the form

$$a_1x + b_1y + c_1$$ = 0

$$a_2x + b_2y + c_2$$ = 0

where,

$$a_1$$ = 2,  $$b_1$$ = 3,  $$c_1$$ = -7

$$a_2$$ = (a – b),  $$b_2$$ = (a + b),  $$c_2$$ = -(3a + b – 2)

For the linear equations to have infinitely many solutions,

$$a_1\over a_2$$ = $$b_1\over b_2$$ = $$c_1\over c_2$$

Here,  $$a_1\over a_2$$ = $$2\over a – b$$,  $$b_1\over b_2$$ = $$3\over a + b$$,  $$c_1\over c_2$$ = $$-7\over -(3a + b – 2)$$ = $$7\over 3a + b – 2$$

$$\implies$$  $$2\over a – b$$ = $$3\over a + b$$ = $$7\over 3a + b – 2$$

$$\implies$$  $$2\over a – b$$ = $$3\over a + b$$   and  $$3\over a + b$$ = $$7\over 3a + b – 2$$

$$\implies$$  2a + 2b = 3a – 3b   and    9a + 3b – 6 = 7a + 7b

$$\implies$$  2a – 3a = -3b – 2b    and     9a – 7a = 7b – 3b + 6

$$\implies$$  a = 5b   ……(1)  and  a = 2b + 3    ….(2)

Solving equation (1) and (2), we get

5b = 2b + 3 $$\implies$$  b = 1

Substituting the value of b in equation (1), we get

a = 5

Thus,  a = 5 and b = 1.

(ii)  The given linear equations can be written as

3x + y – 1 = 0

(2k – 1)x + (k – 1)y – (2k + 1) = 0

Above equations is of the form

$$a_1x + b_1y + c_1$$ = 0

$$a_2x + b_2y + c_2$$ = 0

where,

$$a_1$$ = 3,  $$b_1$$ = 1,  $$c_1$$ = -1

$$a_2$$ = (2k – 1),  $$b_2$$ = (k – 1),  $$c_2$$ = -(2k + 1)

For the linear equations to have no solutions,

$$a_1\over a_2$$ = $$b_1\over b_2$$ $$\ne$$ $$c_1\over c_2$$

$$\implies$$  $$3\over 2k -1$$ = $$1\over k – 1$$   and  $$1\over k – 1$$ $$\ne$$  $$-1\over -(2k + 1$$    and  $$3\over 2k -1$$ $$\ne$$  $$1\over (2k + 1$$

$$\implies$$  3k – 3 = 2k – 1  and   2k + 1 $$\ne$$  k – 1  and  6k + 3 $$\ne$$  2k – 1

$$\implies$$  k = 2   and   k$$\ne$$  -2   and  k $$\ne$$ -1

Hence, the given linear equations has no solutions for k = 2   and   k$$\ne$$  -2   and  k $$\ne$$ -1