Recent content by Mic :)

For h = 2, k = 1/2 and h = -3, k = -1/3, there are infinitely many solutions. For h = 2, k ≠ 1/2 and h = -3 and k ≠ -1/3, there are no solutions. I don't think we've agreed on something for b)

What if h = 2 and k ≠ 1/2. What if h = -3 and k ≠ -1/3? 0= a value Which is not possible.

No, it suggests c) Is there a single definitive equation or is can we just sub any value of h that isn't 2 / -3? eg, 1 -----> x +6y =2 The equations do not result in solutions. For h=0, the hx + 6y =2 gives y=1/3, a unique solution?

If h = 2 and k = 1/2, If h = -3 and k = -1/3 There are no solutions for each case ( 0=0), so they are the values for a)? If h or k = 0, there is one solution (h can be any value if k is 0?); b)? Do we have c) when the left side = 0 or simply for h ≠ 2 and h ≠ -3?

Do we use x + (h+1) = 2k to find infinitely many? when h = -1 and k = any real no.?

Sub in 2 / -3 into the equation, Find k. k = 0.5 for 2 k= -1/3 for -3 So those are the values of h and k for which there are no solutions? How would I go about this for a unique solution / infinitely many?

h=2, -3? Sub the values into 1, h+1 II 2k? Sorry about this, my head's not functioning.

Not if h = 3 or -2?

(h^2 + h - 6)y = 2kh -2?

When the x coeff. is 0, the y coeff. (h^2 + h - 6) is = to 2kh -2? Probably not but I seem a little lost as to what needs to be done.

Should I attempt to solve 6-h(h+1) = 2-2kh? (I arrived at this earlier)

So I have swapped the rows and made the bottom left = 0 1, h+1 II 2k 0, 6/h -h -1 II 2/h -2k Am I on the right tracks?

I don't seem to be able to. Is it possible to multiply a line by 1/h?

Add -----> 2x=6 , x=3 Sub. -----> 2y=4, y =2

Is there a way to determine the number of solutions by subtracting one from the other?