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anonymous
 one year ago
(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
anonymous
 one year ago
(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

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yolomcswagginsggg
 one year ago
Best ResponseYou've already chosen the best response.0Let'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

yolomcswagginsggg
 one year ago
Best ResponseYou've already chosen the best response.0thats for the first equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh.. thank you!! that was easier than i expected

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[6x  2y = 2 \\2 + 6x = y\] you can substitute \(2+6x\) for \(y\) in the first equation, since that is what it is you get \[6x2(2+6x)=2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@pokiedokie it might seem easy but that answer is totally wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you understand what i wrote above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i do understand a bit of it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you solve \[6x2(2+6x)=2\] for \(x\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no it is not arithmetic, it is algebra first use the distributive law to get rid of the parentheses

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH, okay thats what i meant sorry, yikes... so i distributed, is this correct: 6x  4 + 12x = 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now combine like terms on the left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry, no it is not right, but it is close

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[6x  4 12x = 2 \] is more like it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay, give me a second to combine

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i'm terrible at math but do you mean combining 6x with 12x or 4 and 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean do \[6x12x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im pretty sure the placement in what i put is wrong but it is x  46x = 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so \[6x12x=6x\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you should have \[6x4=2\] add \(4\) next

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then i just divide

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 the answer is 1, correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much, sorry for being a pain in the butt!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no problem btw this makes it "consistent" the fact that you get an answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0noted, thank you very much!
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