A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Systems of Equations, please help =)
Solve without matrix
5xy+z=4; x+2y2=5; 2x+3y3=5
I have the answer, but have tried multiple times and cannot get to it.
anonymous
 one year ago
Systems of Equations, please help =) Solve without matrix 5xy+z=4; x+2y2=5; 2x+3y3=5 I have the answer, but have tried multiple times and cannot get to it.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0WAIT that's the wrong equation!! It's 5xy+Z=4, x+2yZ=5, 2x+3y+3Z=5

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so which letter do you want to eliminate... y or z..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whichever is easier.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1ok... so we'll eliminate z you'll end up with 2 equations in 2 unknowns so add equations 1 and 2 5x y + z = 4 x + 2y  z = 5  6x + y = 9 equation 4 now multiply equation 2 by 3 and add equation 3 3x + 6y 3z = 15 2x + 3y + 3z = 5  5x + 9y = 20 equation 5 easiest thing to do now is to multiply equation 4 by 9 and subtract equation 5 which will eliminate y 54x + 9y = 81  5x  9y = 20  49x = 61

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1you'll need to check the calculations

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the unlucky thing about the question you posted is that the answers are all ugly fractions... but I hope it helps you should be able to get x then substitute it into either equation 4 or 5 to find y then substitute the values of x and y into any of the original equations to find x hope it makes sense

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1that's x... very ugly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Huh. The answer is x=1, y=3, z=2...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well you need to check the equations 1st seems ok... 5  3 + 2 = 4 2nd 1 + 6  2 = 5 3rd 2 + 9 + 6 = 17 you have 5 can you check the 3rd equation in the repost....

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so it looks like the last equation should be 2x + 3y  3z = 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. Oops!! So sorry

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1ok... so equation 4 is fine... now multiply equation 1 by 3 and then add equation 3\\ 15x  3y + 3z = 12 2x + 3y + 3z = 5  17x = 17 so now that makes sense... find x, substitute it into equation 4 to find y then substitute the values into any of the original equations to find z hope it now makes sense.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1oops should read 3z = 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes!! thank you so much!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.