Question

PLAESE HELP THIS IS DUE TODAY!!!!!!!
a system of inequalities is shown below
8x+5y=9
3x+2y=4
part A. create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. show the steps to do this.
part b. show the equivalent system of equation has the same solution as the original system of equations,

Answer 1

First, it tells you to replace the 2 equations with the new equations, which are created by adding the 2 original equations, and multiplying one of the original equations by a constant.

Answer 2

Can you do the first part?

Answer 3

im really bad at math so I don't think I can

Answer 4

First of all, can you add the 2 equations?

Answer 5

8x+5y=9
3x+2y=4
---------
11x+7y=13

Answer 6

this ?

Answer 7

you don't necessarily want to straight up add the two equations together. it doesn't get you any closer to the solution.

instead, decide on a variable you'd like to get rid of. either x or y. and which equation you want it removed from.

Answer 8

okay. so the 1st eq. is 8x + 5y = 9. now, you only have the 2nd equation to use to get rid of the y in the first eq.

the 2nd eq. is 3x + 2y = 4. looking at the y terms only, what is the least common multiple of the y coefficients?

Answer 9

so the least common multiply of 5 and 2 is 12?

Answer 10

very good. okay, so now we want the coefficient on the y terms to be 10 and -10 so when we add the 2 equations together, the y term will drop out.

what can i multiply the 1st equation by to get its y term to be 10?
what can i multiply the 2nd equation by to get its y term to be -10?

Answer 11

not quite... look again and use your noodle.

Answer 12

very close... just missing one thing

Answer 13

well, let's look.

if we mult the 1st eq by 2 we get 2(8x + 5y = 9) => 16x + 10y = 18

and if we mult the 2nd eq by 5 we get 5(3x + 2y = 4) => 15x + 10y = 20

10y + 10y \(\ne\) 0

Answer 14

no, look at the posts again. use your noodle and pay attention to the goal and how we planned on acheiving it.

Answer 15

what did we want to do?

Answer 16

just any old equation?

Answer 17

please look at the posts. then you should be a bit clearer

Answer 18

yes, but what were the specifics our goal? our plan?

we had a dialog, stating what we would do and how we'd get there. the point you're missing is trivial, but you're missing the whole idea, in a way.

whenever you get stuck or can't remember what your we're doing and why, you should go back and look at how you arrived at the point where you got stuck, keeping in mind the purpose of what you were doing.

Answer 19

how were we going to do that? the idea is that you learn how to do this because at some point you'll need to be able to do this on your own.

Answer 20

im trying to learn but im on a deadline soooo I need the fast version and main points of the answering process... I already spent an hour on this one question and haven't got anywhere..

Answer 21

from a previous post:
"very good. okay, so now we want the coefficient on the y terms to be 10 and -10 so when we add the 2 equations together, the y term will drop out.

what can i multiply the 1st equation by to get its y term to be 10?
what can i multiply the 2nd equation by to get its y term to be -10?"

Answer 22

nope

Answer 23

http://www.purplemath.com/modules/systlin1.htm

Answer 24

no, i said you were close.

Answer 25

2(8x + 5y = 9) => 16x + 10y = 18
5(3x + 2y = 4) => 15x + 10y = 20

add them together and you get

31x + 20y = 38. the y term remains

Answer 26

pgpilot326
Medals 0

from a previous post: "very good. okay, so now we want the coefficient on the y terms to be 10 and -10 so when we add the 2 equations together, the y term will drop out. what can i multiply the 1st equation by to get its y term to be 10? what can i multiply the 2nd equation by to get its y term to be -10?"

Answer 27

See? Can you do the rest?

Answer 28

is this part a?