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anonymous
 4 years ago
solve the system: y= 1/3 + 2 and x + 3y = 3
anonymous
 4 years ago
solve the system: y= 1/3 + 2 and x + 3y = 3

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x+3y=3 substrct x from both sides what would you get?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0id set them both equal to y then solve like that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ar you can sub y=1/3x+2 in to other equation x+3(1/3 x+x)=3 solve for x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since y equals 1/3x+2 we can substitute that into the second equation like this:\[x+3(1/3 x+2)=3\]and then solve for x:\[x3x+6=3\]\[2x+6=3\]\[2x=3\]\[x=3/2\] now substitute this x into the first equation:\[y=1/3(3/2)+2\]\[y=1/2+2\]\[y=5/2\] \[{x=3/2~~~~~~~~y=5/2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay I see the first part. but the problem with 5/2, is that thats not one of the answer options. but 3/2, 0 is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x+3(−1/3x+2)=3\] and then solve for x: \[x−x+6=3\]\[0x+6=3\]\[0x=−3\]\[x=0\] now substitute this x into the first equation: \[y=−1/3(0)+2\]\[y=0+2\]\[y=2\] \[x=0~~~y=2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is that one of the choices? sorry i did that wrong but i fixed it...helpful? :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(0, 1) (1, 0) (3, 1/3) (3/2, 0) these are the options.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y has to equal 1 when x is 0

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.0yummydum, if 0x=−3, then 0 = 3 which is a contradiction So there are no solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0besides the mistake then you .. wait if you have choses then the order pair are (x,y) sub into the equations example take the first one 0,1 y= 1/3 + 2 and x + 3y = 3 1=1/3(0)+2 and 0+3(1)=3 1=2 not a solutation and 3=3 solves right equations so do the next set of pairs and tell me what you find out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is that all the possiablities? there is no solutions be these are parallal lines

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry your confused but it says solve so look at what we have sub the points in the equations and see if it equals or solve for x or y and then find what you did nt solve for... do you understand this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is a no solution

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.0i would be confused too...the answer choices are incorrect because the true answer is "no solution" or "there is/are no solution(s)"

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.0so there must be a typo somewhere

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is your answer but do you understand why?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think i got this jim_thompson5910

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0great you also can see let equations equal 1/3 x+2=1/3 x+1and solve for x then you get 2=1 so no solution. :)

phi
 4 years ago
Best ResponseYou've already chosen the best response.0another way to look at \[y= \frac{1}{3}x + 2 \text { and } x + 3y = 3\] the 2nd equation, after rearranging into y= mx+b form is \[ y= \frac{1}{3}x+3\] You have two lines that are parallel, and never meet. There is no (x,y) pair that is on both lines (as would be the case if they intersected)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0already said this phi
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