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Mathematics 8 Online
OpenStudy (anonymous):

The table below shows four systems of equations: (System 1) 4x − 5y = 2 3x − y = 8 (System 2) 4x − 5y = 2 10x − 7y = 18 (System 3) 4x − 5y = 2 3x − 8y = 4 (System 4) 4x − 5y = 2 10x + 3y = 15 Which pair of systems will have the same solution? A.) System 1 and system 2, because the second equation in system 2 is obtained by adding the first equation in system 1 to two times the second equation in system 1 B.) System 2 and system 3, because the second equation in system 3 is obtained by adding the first equation in system 2 to two times the second equation in system 2 C.) System 1 and system 2, because the second equation in system 2 is obtained by adding the first equation in system 1 to three times the second equation in system 1 D.) System 2 and system 3, because the second equation in system 3 is obtained by adding the first equation in system 2 to three times the second equation in system 2

OpenStudy (anonymous):

help? @BeautifulMystery. @mathwizzard3 @parker.goodbar @pooja195 @readergirl12

OpenStudy (anonymous):

@Nefxrious What is so hard about this? You can just test all the answers out

OpenStudy (anonymous):

@Needhelpstudying I tried but I don't understand what they are telling me

OpenStudy (anonymous):

@Nefxrious Alright, hold on for a bit, @mathmale is helping me with another question. I'll help you in a bit.

OpenStudy (anonymous):

@Needhelpstudying ok!

OpenStudy (mathmale):

Hello, Neffy. Pairs of different equations can sometimes have the same solution. You may have to solve all four given systems of linear equations, fnding a solution for each. Are any 2 of these four solutions the same? If so, you've answered this question.

OpenStudy (mathmale):

I'm sorry, but need to attend to other things right now. I do urge you to solve all 4 systems of linear equations given you in this problem, and then compare the solutions to see whether or not any 2 are the same.

OpenStudy (photon336):

@Nefxrious do you know how to solve systems of equations?

OpenStudy (anonymous):

@Photon336 Yes

OpenStudy (photon336):

Yeah, it's probably just easier to do what @mathmale said and just solve them one by one. we can start by working on the first one

OpenStudy (anonymous):

@Photon336 I did that and I got all of them wrong except one but it isn't an answer

OpenStudy (photon336):

sometimes it's better to eliminate one variable and then combine the two equations rather than doing substitution. to get rid of y multiply one by 3 the other by -5 3(4x − 5y = 2) 5(10x + 3y = 15) then combine 12x-15y =6 50x+15y =75 62x = 81 x = 1.31~ 4x-5y = 2 4(1.31)-5y = 2 5.24-5y = 2 y = 0.646 x = 1.31, y = 0.646 4x-5y = 2 4(1.31)-5(0.646) = 2.01

OpenStudy (photon336):

eliminating a variable method is sometimes easier than substitution.

OpenStudy (anonymous):

@Photon336 how did you get 3(4x − 5y = 2) 5(10x + 3y = 15)?

OpenStudy (photon336):

so do you notice something? I wanted to get rid of y so i multiplied one equation by 3 and the other by 5 and heres what i got 12x-15y = 6 + 50x+15y = 75 when I add the top and bottom notice something -15y and +15y cancel out. so we're left with 62x = 81 so I multiplied both the top and equation by those numbers.

OpenStudy (anonymous):

@Photon336 where did you get 3 and 5 ?

OpenStudy (photon336):

@Nefxrious that wasn't in the problem originally I multiplied the equations by 3 and 5 so I could get rid of y when I combined them.

OpenStudy (anonymous):

@Photon336 Oh okay! So I just need to do that for the other equations?

OpenStudy (photon336):

yeah, but that's just for that particular one think to yourself: what numbers would you need to multiply the equations by to eliminate a variable.

OpenStudy (photon336):

you get what I did right? @Nefxrious ?

OpenStudy (anonymous):

@Photon336 Yeah! I get! My big problem was trying to cancel out the variable.

OpenStudy (photon336):

the point is canceling out the variables

OpenStudy (photon336):

try this one 4x − 5y = 2 3x − y = 8

OpenStudy (photon336):

ask yourself? what variable do you want to eliminate: pick one

OpenStudy (anonymous):

Photon336 I want to eliminate y

OpenStudy (photon336):

okay so We can eliminate this by multiplying the bottom equation by -5 check this out see how this goes. TOP 4x-5y = 2 Bottom -5(3x-y = 8) TOP 4x-5y = 2 BOTTOM -15x+5y = -40 notice when we add these two equations together -5y and +5y cancel out

OpenStudy (anonymous):

@Photon336 then you would solve and get -11x=-38 -11/-11x=-38/-11 x= 3.45

OpenStudy (photon336):

Then you can plug 3.45 = x back into the top one and you can get y 4x-5y = 2

OpenStudy (anonymous):

@Photon336 4(3.45)-5y=2 13.8-5y=2 13.8-13.8-5y=2-13.8 5y/5=-11.8/5 y=-2.36

OpenStudy (photon336):

Yes nice job y = +2.36 not -2.36 x = 3.45 for this one.

OpenStudy (anonymous):

@Photon336 why not negative? I plugged it in and I got 2.

OpenStudy (photon336):

I did it like this 13.8-2 = 11.8/5y = 2.36 let's check our answers they have to satisfy both the equations or else we're wrong. x = 3.45, y =2.36 4x − 5y = 2 4(3.45)-5(2.36) = 2 correct 3(3.45) − 2.36 = 8 correct we find that these two answers are correct b/c we plugged them into the top and bottom equations and got the answers we were supposed to get here.

OpenStudy (anonymous):

@Photon336 okay I made the 11.8 a negative

OpenStudy (photon336):

hm, okay but for me i just hate using negative numbers lol so when i can i try to make everything positive

OpenStudy (anonymous):

@Photon336 when it looks like this 4x-5y=2 10x-7y=18 would you do -7(4x-5y=2) 5(10x-7y=18)?

OpenStudy (photon336):

Yep you're getting the hang of it when you add those together you'll get -35y and +35y

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