Find the value of m & n for which the system of Linear equation has infinity many solutions. `3x + 4y=12`
`(m+n )x + 2 (m-n)y=5m-1`

Options

A. m = 5 , n = 1
B. m = 5 , n = 2
C. m = 6 , n = 1
D. m = 7 , n = 1

Answer

A

Explanation

The given system of equation is
` 3x + 4y = 12 `,
` ( m +n ) x + 2 ( m - n ) y + ( 1 - 5m ) = 0 `
These equations are of the form
` a_ 1 x + b_ 1 y + c _ 1 =0 and a_ 2 x+ b_ 2 y + c _ 2 = 0 `
where ` a_ 1 = 3, b_ 1 = 4, c_ 1 = - 12 `
and ` a_ 2 = ( m + n ), b_ 2 = 2 ( m -n ), c_ 2 = ( 1 - 5m )`
` therefore (a_ 1 ) /(a_ 2 ) = ( 3)/(( m+ n )' ) (b _ 1 ) /(b _2) = (2 ) /(( m - n )) and (c_ 1 ) /( c_ 2) = ( 12 ) /((5m - 1 )) `
Let the given system of equations have infinitely many solutions.
Then, ` (a_ 1 ) /(a_2 ) = (b_ 1 ) /( b_ 2 ) = (c _ 1) /(c_ 2 ) `
` rArr (3)/(( m+ n )) = ( 2 ) /(( m - n )) = ( 12 ) /(( 5m - 1 ))`
` rArr ( 3)/(( m+ n )) = ( 2 ) /( ( m - n )) and (2 ) /(( m - n )) = ( 12 ) /((5m - 1 ))`
` rArr 3m - 3n = 2m + 2n and 10 m - 2 = 12 m - 12n `
`rArr m - 5n = 0 " " ` ... (i) and ` m - 6n + 1 = 0 " " `... (ii)
On solving (i) and (ii), we get `m = 5 and n = 1 `
Hence, the required values are ` m = 5 and n = 1 `