|
|
Subject:
Geometry
Category: Science > Math Asked by: teatea-ga List Price: $6.00 |
Posted:
05 Dec 2003 04:06 PST
Expires: 04 Jan 2004 04:06 PST Question ID: 283772 |
a. Solve by using substitution or elimination: 3x + 2y = -12 and 2x + y = -7 b. Solve as above: 2x - 4y = -2 and 4x + 8y = 0 c. Solve as above: X - 2y = 9 and 3x + 2y = -5 d. Solve by using substitution and elimination: x = 3y + 11 and 2x + 5y = 0 |
|
Subject:
Re: Geometry
Answered By: endo-ga on 05 Dec 2003 04:34 PST Rated: |
Hi teatea, I will use the same format as last time. Give you a few pointers, then leave lots of space between the answers. a. 3x + 2y = -12 2x + y = -7 Substitution involves expressing one unknown in function of the other unknown. Lets express y in function of x by using the second equation. We have: 3x + 2y = -12 y = -7 - 2x We then replace y in the first expression with the value of y in the second. 3x + 2*(-7 - 2x) = -12 y = -7 - 2x The left expression will give us x, we then replace that value in the right expression. 3x - 14 - 4x = -12 y = -7 - 2x -x = 2 y = -7 + 2*2 x = -2 y = -3 b. In this one we'll use elimination. Elimination works as follows: If you have expressions you can multiply them by real numbers and get an equivalent expression. You can also add two expressions. We have: 2x - 4y = -2 4x + 8y = 0 Multiply the second one by 1/2 2x - 4y = -2 2x + 4y = 0 Now add the right one to the left one. 4x = -2 2x + 4y = 0 x = -1/2 -1 + 4y =0 x = -1/2 y = 1/4 c. x - 2y = 9 3x + 2y = -5 You must have gotten the gist of it now. x = 9 + 2y 3x + 2y = -5 x = 9 + 2y 27 + 6y + 2y = -5 x = 9 + 2y 8y = -32 x = 9 + 2y y = -4 x = 9 - 8 y = -4 x = 1 y = -4 d. x = 3y + 11 and 2x + 5y = 0 substitution: x = 3y + 11 2*(3y+11) + 5y =0 x = 3y + 11 6y + 22 + 5y =0 x = 3y + 11 11y = -22 x = 3y +11 y = -2 x = -6 +11 y = -2 x = 5 y = -2 elimination x = 3y + 11 2x + 5y =0 x - 3y = 11 2x + 5y =0 -2x + 6y = -22 2x + 5y =0 -2x + 6y = -22 11y = -22 -2x + 6y = -22 y = -2 -2x - 12 = -22 y = -2 -2x = -10 y = -2 x = 5 y = -2 I hope this answers your question. If you need any more help or clarifications please do not hesitate to ask. Thanks. endo |
teatea-ga rated this answer: |
|
There are no comments at this time. |
If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you. |
Search Google Answers for |
Google Home - Answers FAQ - Terms of Service - Privacy Policy |