Ok, to find a solution you plug in a RANDOM value in for x and see what comes out for y.
Let's pick these values for x:
x = 1
x = 2
x = 3
Now we plug them in and find the y values:
7y + 1 = 21
First we subtract one from both sides:
7y = 20
Then we divide both sides by 7
y = 20/7
y = 2.9
That is one solution.
7y + 2 = 21
We do the same thing as before
7y = 19
And the next step is also the same:
y = 19/7
y = 2.7
The Second Solution.
7y + 3 = 21
7y = 18
y = 18/7
y = 2.6
And there's your third solution.
2007-04-14 14:59:55
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answer #1
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answered by Eolian 4
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Well if x + 7y = 21 an obvious solution is
x = 14, y =1.
So we can get two more solutions by decreasing x:
x = 7 y = 2
x = 0 y = 3.
When x decreases by 7, y increases by 1.
That's because y = -x/7 + 3 and the slope
of this line is -1/7.
2007-04-14 15:09:59
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answer #2
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answered by steiner1745 7
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1.y=2 and x=7
2.y=4 and x= -7
3.y=5 and x= -14
1. 7(2)+7=21
14+7= 21
21=21
2. 7(4)+(-7) = 21
28-7=21
21=21
3. 7(5)+(-14) = 21
35-14=21
21=21
but really the possibilites are endless all you have to do is pick a number for 'y' then multiply by 7 and then subtract 21 from that and the answer would be your 'x' value.
2007-04-14 15:00:39
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answer #3
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answered by CreamSaver2 2
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i will purely think of of three a threat the right thank you to do it: a million. 4 packing containers of 20 2. 5 packing containers of 8 and a couple of packing containers of 20 3. 10 packing containers of 8 i don't be attentive to if there is any artwork to tutor cuz i only kinda did it in my head by using fixing the equation 8x + 20y = 80 so i think of there is entirely some wager artwork in touch however the 4 packing containers of 20 is extra low-priced do only the math you do 4*7 to be sure how plenty option a million expenses and 5*3.12 + 2*7 to be sure how plenty option 2 expenses and 10*3.12 for option 3
2016-10-22 04:43:41
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answer #4
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answered by tonini 4
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7y+x=21
If y = 1
7(1)+x=21
7 + x= 21
x = 21-7
x = 14
If y = 3
7(3)+x=21
21+x=21
x = 21-21
x= 0
If y = 0
7(0)+x=21
0+x=21
x=21
2007-04-14 15:03:39
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answer #5
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answered by Anonymous
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7x3=21 + 0 = 21
7x2=14 + 7= 21
7x0=0 + 21 = 21
7x 1=7 + 14= 21
7x.5 =3.5 + 17.5 = 21
and the beat goes on.............
2007-04-14 15:41:40
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answer #6
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answered by ♣Hey jude♣ 5
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7y+x=21
7y=21-x
y=3-(1/7)x
just plug any number in as x and you can get answers for y.
2007-04-14 14:55:33
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answer #7
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answered by fuzzynaval 1
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7y+x=21
7y=21-x
7y-21=-x
-7y+21=x
7y+(-7y+21)=21
y=0
7(0) + x = 21
therefore: x=21; y=0
2007-04-14 15:00:12
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answer #8
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answered by Michelle Moy 2
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(21,0), (0,3), (14,1) are three solutions.
How did I get them,?
#1 - let x =0, solve for y 7y=21, y=3 (0,3)
#2 - let y = 0, solve for x x=21 (21,0)
#3 - let y = 1, solve for x 7 + x = 21, x=14 (14,1)
2007-04-14 14:55:59
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answer #9
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answered by bz2hcy 3
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hm.....
if you graph this equaton... you can get an infinite number of solutions
just oberve...
let y=0.. x has an answer of 21
let y=1.. x has an answer of 14
let y=2.. x has an answer of 7
let y=3.. x has an answer of 0
let y=4.. x has an answer of -7
let y=5.. x has an answer of -14
let y=6 ..............
it is infinite.
you need to provide at least one more equation...
2007-04-14 15:02:38
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answer #10
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answered by ken_dicey 2
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