Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Choose h and k such that the system has (a) no solution (b) a unique solution or (c) many solutions. Give separate answers for each part. x1 - 3x2 = 1 and 2x1 + hx2 = k

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Still trying to get through these review questions from what I missed while ill.

- anonymous

\[x_1-3x_2=1\\
2x_1+hx_2=k\]?

- anonymous

Yep, those are the correct systems. I should have made sure to mention that they were row labels and not exponents. I'm sorry if that was confusing.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

i got it

- anonymous

if you pick some random \(h\) and \(k\) there will probably be one unique solution

- anonymous

if you multiply the first equation by 2 you get
\[2x_1-6x_2=2\\
2x_2+hx_2=k\]

- anonymous

if you make \(h=-6\) and \(k=2\) they will be the same line exactly so there will be infinitely many solutions

- anonymous

if you make \(h=-6\) and \(k=\) anything but 2, they will be parallel lines with no intersection

- anonymous

Oh wow! How did you figure that out?

- anonymous

i am the satellite

- anonymous

ok in truth if you wanted to solve it you would have to match up the coefficients to use "elimination"
i just matched up the coefficients of the \(x_1\) term by doubling the first one

- anonymous

Hehehe, that makes sense.

- anonymous

Thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.