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Find conditions on k that will make the following system of equations have a unique solution. To enter your answer, first select whether k should be equal or not equal to specific values, then enter a value or a list of values separated by commas.Then give a formula in terms of k for the solution to the system, when it exists. Be sure to include parentheses where necessary, e.g. to distinguish 1/(2k) from 1/2k.3kx+3y = 43x+3ky = 1The system has a unique solution when k (equal or not equal) to ???The unique solution is x = y =

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shejladautovic

Answer:

[tex]k\not = -1,1[/tex]

[tex]x=\frac{4k-1}{3(k^2-1)}[/tex]

[tex]y=\frac{k-4}{3(k^2-1)}[/tex]

Step-by-step explanation:

First we see that 3k/3 can't be equal to 3/3k, so we have

[tex]9k^2\not =9[/tex], i.e, k can't be -1 or 1.

multiple first equation with k:

[tex]3k^2x+3ky=4k[/tex]

[tex]3x+3ky=1[/tex]

then we have:

[tex]3(k^2-1)x=4k-1[/tex]

i.e,

[tex]x=\frac{4k-1}{3(k^2-1)}[/tex]

And now we calc y:

[tex]3k\frac{4k-1}{3(k^2-1)}+3y=4[/tex]

[tex]y=\frac{k-4}{3(k^2-1)}[/tex]

for any k not eql to -1 or 1.

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