The answer is pay attention in school more and do your own homework. Why aren't your parents helping you?
2007-12-14 05:22:39
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answer #1
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answered by Sapphire 5
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2x + 4y = 16 (1)
3x - y = 3 (2)
multiply (2) by 4
12x - 4y = 12 (3)
add (1) and (3)
2x + 4y = 16 (1)
12x - 4y = 12 (3)
14x = 28
x= 28/14 = 2
substitute x = 2 in (2)
3(2) - y = 3
6 - y = 3
- y = 3 - 6
- y = -3
y = 3
therefore x = 2 y = 3
2) x - 2x - 3 = 3
- x -3 = 3
-x = 3+3
- x = 6
x = -6
~~~
2007-12-14 05:24:52
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answer #2
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answered by A Little Sarcasm Helps 5
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For number 1:
isolate the y in the first equation by subtracting the 2x so you get 4y=16-2x....then divide both sides by 4 to isolate the y....y=4-.5x...plug what y equals into the second equation (3x-y=3) so you get 3x-4-.5x=3. Combine the two x's: 2.5x-4=3...add 4 to both sides....2.5x=7...divide by 2.5 to isolate the x....x=2.8....take the x=2.8 and plug it into the first equation of y=4-.5x so you get y=4-.5 (2.8)....y=4-1.4...y= 2.6.
X=2.8
Y=2.6
For number 2:
x-2x-3=3
first combine the x's to get -x-3=3...add 3 to isolate the x....-x=6 divide by -1, x=-6
X=-6
2007-12-14 05:33:35
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answer #3
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answered by jalamagirl 2
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1. x= (16 - 4y)/ 1.2
Replace the x in 3x-y=3
3[(16 -4y)/1.2] - y=3
40 -10y -y=3
-11y= -37
y= 37/11
replace y in x
x=70/33
2. x-2x=3+3
-x=6
x=-6
2007-12-14 05:30:28
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answer #4
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answered by MG 2
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2x+4y=16
3x-y=3
-y=3-3x
-y/-1=3-3x
y=3x-3
2x +4(3x-3)=16
2x+12x-12=16+12
14x=28
x=2
3(2)-y=3
6-y=3
y=3
x-2x-3=3
-x-3=3
-x=-6
x=6
2007-12-14 06:03:31
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answer #5
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answered by Dave aka Spider Monkey 7
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Eh... neither... yet i assume i'm going to come to a decision on English. I used to love English extra yet intense college made it uninteresting. -_- yet I nonetheless love writing as a interest (and hate math, even however i'm no longer undesirable at it) so i'm going to basically say English.
2016-10-11 06:59:20
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answer #6
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answered by figurelli 4
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is it a simultanious equation??
if it it, 1. 2x + 4y
12x - 4y
14x = 12
x=0.86
2007-12-14 05:27:11
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answer #7
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answered by Anonymous
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ok..u got the answer now..who would help u in ur exams ?
2007-12-14 05:28:02
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answer #8
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answered by Sid 3
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