For what values of m and n the following system of linear equations ha

1.	For what values of m and n the following system of linear equations has infinitelymany solutions.3x+4y = 12(n + n) x + 2 (m-n) y = 5m-1
Answer» To have infinite solutions both equation should be same (m+n)x + 2(m-n)y = 5m-1should be equivalent to 3x + 4y = 12 => k3x + k4y = k12 where k is any variable => (m+n)/3 = 2(m-n)/4 = (5m-1)/12 = k Multiplying each with 12=> 4m + 4n = 6m - 6n = 5m -1 = 12k taking 1st 2 equations 4m + 4n = 6m - 6n=> 10n = 2m=> m = 5n - eq A taking 1st & 3rd4m + 4n = 5m-1=> m = 4n + 1 - eq B equating value of m from eq A & eq B5n = 4n + 1=> n = 1 putting value of n in eq Am = 51 = 5 Hence value of m = 5 & n = 1

For what values of m and n the following system of linear equations has infinitelymany solutions.3x+4y = 12(n + n) x + 2 (m-n) y = 5m-1

Answer»

To have infinite solutions both equation should be same

(m+n)x + 2(m-n)y = 5m-1should be equivalent to 3x + 4y = 12

=> k*3x + k*4y = k*12 where k is any variable

=> (m+n)/3 = 2(m-n)/4 = (5m-1)/12 = k

Multiplying each with 12=> 4m + 4n = 6m - 6n = 5m -1 = 12k

taking 1st 2 equations 4m + 4n = 6m - 6n=> 10n = 2m=> m = 5n - eq A

taking 1st & 3rd4m + 4n = 5m-1=> m = 4n + 1 - eq B

equating value of m from eq A & eq B5n = 4n + 1=> n = 1

putting value of n in eq Am = 5*1 = 5

Hence value of m = 5 & n = 1