Question

Help! Will Medal!! :)
A system of equations is shown below:

5x - 5y = 10
3x - 2y = 2

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 that the equivalent system has the same solution as the original system of equations.

Answer 1

I think I've figured out the some of the first part but I don't know where to go from there..

Answer 2

You could subsitute\[3x-2y=2\]\[-2y -3x=2-3x\]\[-2y=2-3x\]\[-2y/-2=(2-3x)/-2\]\[y=(2-3x)/-2\]

Answer 3

Do you know what to do next?

Answer 4

No I'm confused. Could you write out the steps please I'm not sure how you got to each :(

Answer 5

I did! I was just setting one of the equations equal to y

Answer 6

I am not finished yet. I was seeing if you knew where to go from where I left off.

Answer 7

I meant worded form. Like what did you sub in for that equation? And is your goal to isolate y?

Answer 8

I isolated y first

Answer 9

Do you remember how I got\[(2-3x)/-2\]For y?

Answer 10

Well now since I know what y equals I can substitute y like this:
\[ 5x - 5y = 10\] \[5x - 5((2-3x)/-2) = 10\]

Answer 11

Are you following?

Answer 12

Yes.

Answer 13

Alright, now we solve for x.

Answer 14

Okay. Wait is this still only PART A?

Answer 15

I think so

Answer 16

Okay. Would my answer be 6/5?

Answer 17

For x

Answer 18

Oh wait! I'm approaching A the wrong way.

Answer 19

Sorry, it's like this because it is "Replace one equation with the sum of that equation and a multiple of the other."

Answer 20

Nevermind

Answer 21

:( Im confused

Answer 22

Here is an archived question that might be able to help:
http://openstudy.com/study#/updates/52eb025ce4b0d95ca6b33161

Answer 23

This method is what I was doing before and I got:
Solve the system by multiplying the top by 3 and the bottom by -5.
So 15x-15y=30
-15x+10y=-10
This equals -5y=20 and so y=-4

Answer 24

But I don't know if it's right or where I go from here :L

Answer 25

Don't you have to multiply one by a negative though? Otherwise it doesn't cancel out?

Answer 26

no need to as we are just isolating the y and disposing the x.

Answer 27

give us a sec to alter it... jst a slight mistake

Answer 28

Oh okay

Answer 29

I'll write on on paper and scan it and post it.

Answer 30

Okay!

Answer 31

:)

Answer 32

sorry..the handwriting is bad

Answer 33

It's lovely. And thank you. Part B? You might only have to explain it to me because I just don't understand what the question is asking of me.

Answer 34

Just show that that your first pair of equations and your second pair of equations have the same solution by solving for y and x on both sets.

Answer 35

I am not really sure as well. It could be that you have to substitute both "X" and "Y" into both 1a and 2a to prove that they have the same answer. So, you should get the same answers as the first pair of equations using the second pair of equations.

Answer 36

Okay Thanks! I will fan you both

Answer 37

anymore question?

Answer 38

No but I'll let you know if I do! Thanks!