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Algebra 17 Online
OpenStudy (pokiedokie):

(please walk me through this!) Without graphing, classify each system of equations as independent, dependent, or inconsistent. Solve independent systems by graphing. 6x - 2y = 2 2 + 6x = y

OpenStudy (yolomcswagginsggg):

Let's solve for x. 6x−2y=2 Step 1: Add 2y to both sides. 6x−2y+2y=2+2y 6x=2y+2 Step 2: Divide both sides by 6. 6x 6 = 2y+2 6 x= 1 3 y+ 1 3 Answer: x= 1 3 y+ 1 3

OpenStudy (yolomcswagginsggg):

thats for the first equation.

OpenStudy (yolomcswagginsggg):

1/3 y+ 1/3

OpenStudy (anonymous):

yikes

OpenStudy (pokiedokie):

oh.. thank you!! that was easier than i expected

OpenStudy (yolomcswagginsggg):

lol

OpenStudy (anonymous):

\[6x - 2y = 2 \\2 + 6x = y\] you can substitute \(2+6x\) for \(y\) in the first equation, since that is what it is you get \[6x-2(2+6x)=2\]

OpenStudy (anonymous):

@pokiedokie it might seem easy but that answer is totally wrong

OpenStudy (pokiedokie):

oh whoops...

OpenStudy (anonymous):

you are solving a system of equations, i.e. where the lines intersect. you are not solving for x and y in one equation which is impossible

OpenStudy (anonymous):

do you understand what i wrote above?

OpenStudy (pokiedokie):

yes i do understand a bit of it

OpenStudy (anonymous):

can you solve \[6x-2(2+6x)=2\] for \(x\)?

OpenStudy (pokiedokie):

i think so, but i'm still a little confused on where to start in the equation. like as in, do i take care of the parentheses first like in PEMDAS

OpenStudy (anonymous):

no it is not arithmetic, it is algebra first use the distributive law to get rid of the parentheses

OpenStudy (pokiedokie):

OH, okay thats what i meant sorry, yikes... so i distributed, is this correct: 6x - 4 + 12x = 2

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

now combine like terms on the left

OpenStudy (anonymous):

oh sorry, no it is not right, but it is close

OpenStudy (anonymous):

\[6x - 4 -12x = 2 \] is more like it

OpenStudy (pokiedokie):

oh okay, give me a second to combine

OpenStudy (pokiedokie):

sorry, i'm terrible at math but do you mean combining 6x with 12x or 4 and 2

OpenStudy (anonymous):

i mean do \[6x-12x\]

OpenStudy (pokiedokie):

im pretty sure the placement in what i put is wrong but it is x - 4-6x = 2

OpenStudy (anonymous):

what is \(6-12\)?

OpenStudy (pokiedokie):

-6

OpenStudy (anonymous):

ok so \[6x-12x=-6x\] right?

OpenStudy (yolomcswagginsggg):

:P

OpenStudy (anonymous):

so you should have \[-6x-4=2\] add \(4\) next

OpenStudy (pokiedokie):

-6x = 6??

OpenStudy (pokiedokie):

and then i just divide

OpenStudy (pokiedokie):

@satellite73 the answer is -1, correct?

OpenStudy (anonymous):

yes

OpenStudy (pokiedokie):

thank you so much, sorry for being a pain in the butt!!

OpenStudy (anonymous):

no problem btw this makes it "consistent" the fact that you get an answer

OpenStudy (pokiedokie):

noted, thank you very much!

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