Question

For which value of a, does the following pair of linear equations ax - y = 2 and 6x - 2y = 4 have
infinitely many solutions.
(a) 3
(b) -3
(C) 4
(d) -12
​

Answer 1

qυєsтIση :-

• For which value of a, does the following pair of linear equations ax - y = 2 and 6x - 2y = 4 have
• infinitely many solutions.
• (a) 3
• (b) -3
• (c) 4
• (d) -12

αηsωєя :-

• ax - y = 2
• 6x - 2y = 4

• for infinite many solutions
• $\mathtt{ \frac{a _{1} }{a_{2} } = \frac{b_{1} }{b_{2} } = \frac{c_{1} }{c_{2} } }$
• $\mathtt{ \frac{a}{6} = \frac{ - 1}{ - 2 } = \frac{2}{4} }$
• $\mathtt{ \frac{a}{6} = \frac{1}{2 } }$
• $\mathtt{2a = 6}$
• $\mathtt{a = \frac{6}{2} }$
• $\boxed{\mathtt{a = 3}}$
• (a) 3

Answer 2

Given-

A pair of linear equations, x - y = 2 and 6x - 2y = 4

To Find-

The value of 'a' for which the given pair of equations have infinitely many solutions.

Solution-

The condition for infinitely many solutions is-

$\frac{a1}{a2}=\frac{b1}{b2}=\frac{c1}{c2}$    ____(i)

For the given pair of equations, ax - y = 2 and 6x - 2y = 4

a₁= a; a₂= 6

b₁= -1; b₂= -2

c₁= -2; c₂= -4

Putting these values in equation(i), we get

$\frac{a}{6}=\frac{-1}{-2} =\frac{-2}{-4}$

⇒ -2a = -6

⇒ a = 3

Hence, for a = 3, x - y = 2 and 6x - 2y = 4 have  infinitely many solutions.