I've calculated the gradient and k = 2 and -2/3
When k = 2,
2x+3y=8
4x+6y=4
When k = -2/3
-2x+y=24
4x-2y=4
The equations are all similar except for the constants and I do not know which is no solutions or infinitely many solutions.
Where do I go from here? Also how do i find the unique solutions? Thx a bunch
Find the value(s) of k for which the pair of equations:
Equation 1: kx + (k+1)y = 8
Equation 2: 4x + 3ky = 4
(a) no solutions
(b) a unique solution
(c) infinitely many solutions
Can you please use the gradient way and show your working? Thanks a lot
2 answers
Suppose the slopes are equal:
-k/(k+1) = -4/(3k)
-3k^2 = -4k-4
3k^2-4k-4 = 0
(3k+2)(k-2) = 0
k = 2 or -2/3
as you calculated. Using you pairs of equations, we see that they can be written
4x+6y=16
4x+6y=4
2x-y = -24
2x-y = 2
In both cases the lines are distinct and parallel, so there is no solution.
Any other value of k will give the lines different slopes, so they must intersect at one point.
So, there is no value of k which will make the two lines the same, so (c) cannot be obtained.
-k/(k+1) = -4/(3k)
-3k^2 = -4k-4
3k^2-4k-4 = 0
(3k+2)(k-2) = 0
k = 2 or -2/3
as you calculated. Using you pairs of equations, we see that they can be written
4x+6y=16
4x+6y=4
2x-y = -24
2x-y = 2
In both cases the lines are distinct and parallel, so there is no solution.
Any other value of k will give the lines different slopes, so they must intersect at one point.
So, there is no value of k which will make the two lines the same, so (c) cannot be obtained.
1 answer