Find k if the following system of equations has a
unique solution?
2x + (k-1)y = 6
3x+(2k+1)y=9
find k if the following system of equations has
infinitely many solutions?
kx+3y=k-3
12x+ky=k
I have tried these problems but I'm lost I do
understand what a unique and infinite is just
don't understand how to do the problem?
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"Unique" means "one and only one". "infinite" means
there is no end to the number of possible solutions
I disagree with your teacher. For
instance, in the first equation, I think your
teacher thinks that if the term in y, namely
(k-1)y = 0, then k will equal 1 and thus the
equation will become:
2x + (1-1)y = 6
2x + 0y = 6
x = 3
And x will have a unique VALUE 3. However,
having a unique VALUE for x does NOT MEAN
that there is a unique SOLUTION to an
equation which contains two variables.
SOLUTIONS to such equations are ordered
pairs of numbers of the form (x,y), such as
(3,6), and NOT just single numbers like 3.
So even if k = 1, the equation
2x + 0y = 6 will not have a unique solution
because
(x,y) = (3,6), (3,4), (3,-7), (3,1),
(3,9999), (3,0), (3,3.1416), (3,-73.617),
are all solutions and there are infinitely
many more ordered pairs with first
coordinate 3 are also solutions.
If I had to guess what your teacher wants
for the first two, I'd guess he/she wants
you to set the coefficient of y equal to 0,
so it would be k=1 for the first one.
For the second one,
3x + (2k + 1)y = 9
you'd set 2k + 1 = 0
2k = -1
k = -1/2
However, this is only a guess as to what your
teacher is thinking. But your teacher is
thinking incorrectly.
As far as the rest of them, k could be chosen
as any number at all, and there would be an
infinite number of solutions. For the third
equation
kx + 3y = k-3
For instance if we arbitrarily choose k = 4
kx + 3y = k-3
4x + 3y = 4-3
4x + 3y = 1
There are infinitely many solutions, such as
(x,y) = (1,-1) (-5,7), (-2,3), (4,-5), (7,-9),
and infinitely many more.
But if we arbitrarily choose k = -7
kx + 3y = k-3
-7x + 3y = -7-3
-7x + 3y = -10
there are also infinitely many solutions. For instance
(x,y) = (1,-1), (-2,-8), (4,6), (7,13), (-5,-15), and
infinitely many more.
No matter what value we choose for k, there will be
infinitely many solutions. So your teacher could NOT
ask you to find "the one and only one value of k which
will cause the equation to have infinitely many
solutions. Because there ISN'T JUST ONE!!
The same is true for the last one:
12x + ky = k
If we arbitrarily choose k = -29.7
12x + ky = k
12x + (-29.7)y = -29.7
12x - 29.7y = -29.7
then we have these solutions:
(x,y) = (0,1), (-5, -1.02), (42, 17.97), (-49,-18.8)
and infinitely many more.
So you can tell your teacher that at least one other
mathematics teacher claims that this assignment is
flawed. ;-) But please don't be arrogant or rude
to him/her!
Edwin
