00:01
Where as to solve the given system of equations and then determine the values of a for which there are no solutions, exactly one solution and infinitely many solutions.
00:12
What we'll do is multiply the second equation by 3 and add to the first equation to get rid of this 3y.
00:23
Y will get negative 2x plus 3ax, 3y will get cancelled and 5 minus 3 equals 2.
00:38
And the second equation is going to be the same.
00:42
Now we'll solve the first equation for a.
00:45
For x we'll get negative 2 plus 3ax equals 2, ax minus y equals 1 and then x equals 2 divided by negative 2 plus 3a and from the second equation y equals ax minus 1 and after plugging in x we'll get y equals a multiplied by 2 over 3a minus 2 minus 1 and this equals 2a over 3a minus 2 minus 3a minus 2 over 3a minus 2 and this equals negative a plus 2 divided by 3a minus 2.
01:46
So this is a solution in terms of a.
01:49
X equals 2 over 3a minus 2 and y equals 2 minus a over 3a minus 2.
02:03
Now we need to determine the values of a for which there are infinitely many solutions, no solutions or the unique solution and to do that we need to get back to this equation here i'll copy it and if we look at that so first for the system to have no solutions we want the expression on the left hand side to be zero in the first equation we'll focus on this equation if negative 2 plus 3a is zero then the expression on the left is zero on the right is non -zero 0 cannot be equal a non -zero number, so there are no solutions...