Ask a New Question

Asked by Hin

The equations
x+ ky + 2z =0
x + (2k-1)y + 3z =0
x + ky + (k+3)z = 2k-1

Find the values of k such that
a) the system has a unique solution

b) the system has no solutions

c) the system has infinitely many solutions
Thanks
9 years ago

Answers

There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!

Submit Your Answer


We prioritize human answers over AI answers.

If you are human, and you can answer this question, please submit your answer.

Related Questions

What are the equations for the following? Positron emission from silicon-26 and S... Which two of the four equations represents parallel lines? These are the other equations I've balanced. I need help with one. AgNO3 + K2Cr2O7 --> double rep... Do the equations x = 5y + 3 and x = 5y – 3 have the same solution? How might you explain your answer... given these equations: x+y=9.0 0.5x+.2y=3.0 find the values of x and y Consider the equations: f(X)= -6x-1 and g(x)=4x^2 Select the solution for (fg)(x). 5) Consider the equations: f(X)= -6x-1 and g(x)=4x^2 Select the solution for (f+g)(x). a) -... Put the following equations in standard form. State the center and the radius. x^2-5x+y^2+4y=-3... Given the following equations, calculate the ΔH of reaction of ethylene gas (C2H4) with fluorine gas... Which of these equations is a non-proportional relationship?(1 point) Responses y=−15x y is equal to...
Submit Your Answer

Question

The equations
x+ ky + 2z =0
x + (2k-1)y + 3z =0
x + ky + (k+3)z = 2k-1

Find the values of k such that
a) the system has a unique solution

b) the system has no solutions

c) the system has infinitely many solutions
Thanks

Ask a New Question
Archives Contact Us Privacy Policy Terms of Use