solve both for y, and you'll have them in y = mx + b notation:
-3y = 9 - 2x
y = 2x/3 - 3
6y = 4x + 8
y = 2x/3 + 4/3
There's your answer. The two lines are parallel, and they will never meet.
2006-12-03 09:06:22
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answer #1
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answered by Dave 6
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If u plot the graphs
for 2x - 3y = 9
u get x = 0, y = -3
x = 2, y = -21/3
x = 3, y = -1
then -4x + 6y = 8
u get x = 0, y = 11/3
x = 1, y = 2
x = 2, y = 22/3
x = 3, y = 4
In simultaneous equations, where the lines cross are the solutions. equations of that form are straight lines.
these lines do not cross. they are parallel and therefore will never meet.
if u check the gradients by differentiation u get
1. 3y = 2x - 9 divide by 3 gives dy/dx = +2/3 (gradient)
2. 6y = 8 + 4x divide by 6 gives dy/dx = +2/3 (gradient)
if the gradients are the same this checks as if they start at different places, which they do, they will never meet.
sure wish i could show u my graphs !!!
i think that's right
2006-12-03 17:26:39
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answer #2
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answered by Anonymous
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2x - 3y = 9 => 2x - 9 = 3y => y = (2/3)x - 3
-4x + 6y = 8 => 6y = 4x + 8 => y = (2/3)x + (4/3)
When the equations are rearranged to the y=mx+c format, you will see that the gradients are the same. This means that the 2 equations are parallel, and hence do not intersect. And so you will not get an answer.
2006-12-03 17:04:42
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answer #3
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answered by Kemmy 6
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It's easy, i'll show you:
(1) : is 2x - 3y = 9
(2) : is -4x + 6y = 8
now i'm dividing (2) by -2:
(2) / -2 = (3) : 2x - 3y = 4
but, how could this be? The left side of (1) and (2) are the same, but the right one are different, so there are no possible x and y to solve both equations. There will always at least one equations be wrong because 4 isn't 9.
2006-12-04 06:07:37
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answer #4
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answered by Anonymous
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Well, I'll say this, when I first started doing these, the first... say... 8 chapters of it were greek, and then it all hit me in the face. Here's how I learned to solve these types of problems:
2x - 3y = 9 *2 -> 4x - 6y = 18 (I multiplied, because one needs
-4x + 6y = 8 *1-> -4x +6y = 1 to be the inverse of the other)
Then we combine the 4x and the -4x, the -6y and the 6y and combine the two equations together to get:
0=19
And the last time I checked my math book zero does NOT equal 19.
If you have any question's I'd be happy to clarify.
2006-12-03 17:05:53
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answer #5
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answered by Anonymous
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Divide the second equation by -2.
You get:
2x - 3y = -4
But you also have
2x - 3y = 9
In order to have any solutions, what is equal to -4 would also have to be equal to 9, an impossibility.
Since these two equations will never both be satisfied by any values of x or y, you have no solutions. In essence the two lines are parallel and will never intersect. In other words, they do not have any points in common (any solutions).
2006-12-03 17:13:04
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answer #6
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answered by Draco Moonbeam 3
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The equations are inconsistent. A trivial algebraic transformation of the second one gives 2x - 3y = -4. Thus, the two equations represent two parallel lines. Only equations having non-parallel lines can have a mutual solution, since the solution is represented by the intersection of the lines.
2006-12-03 17:16:56
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answer #7
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answered by Anonymous
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Try graphing the two linear equations in a coordinate plane and you will arrive at two lines parallel to each other. since there are no intersection, there no common solution to both of them.
Secondly, try using the elimination method and you will arrive at this trivial answer: 0 = 10. Since 0 is not really equal to 10, hence, the two lines have no solution
2006-12-03 17:10:44
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answer #8
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answered by Sheila 2
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The LH side of the equations differ by a factor of -2. As the RH sides clearly don't differ by that factor, there's no solution.
This is not a mathematical solution but the nearest I can get.
2006-12-03 17:13:14
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answer #9
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answered by migdalski 7
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You can't see it easily when they're written that way, but the lines are parallel, but not the same line: hence no intersection.
In slope intecept form the slopes are both 2/3
Putting x (or y) = 0, shows that they have different intecepts, hence parallel but not coincident.
2006-12-03 17:11:35
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answer #10
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answered by modulo_function 7
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