Saved Bookmarks
| 1. |
Find the value of k so that the pair of linear equations x+2y =5 and 3x+ky+5=0 have unique solution |
|
Answer» The given pair of linear equations isx + 2y - 5= 0 and\xa03x + ky + 5 = 0Here, a1\xa0= 1, b1\xa0= 2,a2\xa0= 3, b2\xa0= kFor having a unique solution, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { 3 } \\neq \\frac { 2 } { k } \\Rightarrow k \\neq 6{/tex} Unique solution=a1/a2 not equal to b1/b2here, a1=1; a2=3; b1=2; b2=kthen, 1/3=1/kcross multiplythn k =3 |
|