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its either
(3,9)
(3,-9)
(3, -3)
(3,3)

2007-05-12 12:47:48 · 9 answers · asked by smj 1 in Science & Mathematics Mathematics

9 answers

Solve for "x" or "y" in either equation. Let's solve "y" in the 1st equation > 2x + y = 3

First: subtract 2x from both sides (when you move a term to the opposite side, always use the opposite sign).

2x - 2x+ y = 3 - 2x
y = 3 - 2x

Sec: replace (3 -2x) with "y" in the 2nd equation.

x - 3(3-2x) = 12
x - 3(3) - 3(-2x) = 12
x - 9 + 6x = 12
7x - 9 = 12 > Add 9 from both sides.
7x - 9+9 = 12+9
7x = 21 > Divide both sides by 7.
7x/7 = 21/7
x = 3

Third: replace 3 with "x" in the 1st equation.

2(3) + y = 3
6 + y = 3 > Subtract 6 from both sides.
6-6 + y = 3-6
y = - 3

Solution (3, -3)

2007-05-12 16:35:36 · answer #1 · answered by ♪♥Annie♥♪ 6 · 0 0

2x+y=3 multiply both sides by 3
6x+3y=9
x-3y=12 add the 2 equations
7x=21 divide both sides by 7
x=3
2x+y=3 substitute 3 for x
2*3+y=3
6+y=3 subtract 6 from each side
y=-3
check
x-3y=12 substitute 3 for x & -3 for y
3-3(-3)=12
3+9=12
12=12

solution is (3, -3)

2007-05-12 13:47:49 · answer #2 · answered by yupchagee 7 · 0 0

The answer is (3, -3) assuming it is (X, Y)

2(3) + (-3) = 3 ... 6 - 3 = 3
and
3 - 3(-3) = 12... 3 - (-9) = 12

2007-05-12 13:00:59 · answer #3 · answered by ellusionary 5 · 0 1

(2x+y=3) times 3 equals
6x+3y=9
x-3y=12
-----------+
7x=21
x=3
plug it in

2(3)+y=3
6+y=3
y=-3

2007-05-12 12:55:33 · answer #4 · answered by Anonymous · 0 0

3, -3 is the answer

2007-05-12 12:53:35 · answer #5 · answered by toot 1 · 0 1

6x + 3y = 9
x - 3y = 12-----ADD

7x = 21
x = 3

6 + y = 3
y = - 3

Answer (3, - 3)

2007-05-13 02:51:47 · answer #6 · answered by Como 7 · 0 0

It is a simple one?
how ever it is (3,-3)

2007-05-12 13:26:08 · answer #7 · answered by Anonymous · 0 1

2x+y=3 x-3y=12
x-3y=12
x=12+3y
2x+y=3
2(12+3y)+y=3
24+6y+y=3
7y=3-24
7y=-21
y=-21/7
y=-3
x-3y=12
x-3(-3)=12
x+9=12
x=12-9
x=3

2007-05-12 15:51:30 · answer #8 · answered by sam 3 · 0 0

do your own homework

2007-05-12 12:51:16 · answer #9 · answered by Anonymous · 0 2

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