Can someone help with this elimination question?
回答 (3)
2015-05-24 12:29 am
6x^2 - 3x = 2y - 5 and 2x^2 + x = y - 4. Double the 2nd equation:
4x^2 + 2x = 2y - 8, and subtract that from the 1st equation:
2x^2 - 5x = 3. Collecting terms on 1 side, you have:
2x^2 - 5x - 3 = 0, or
(2x + 1)(x - 3) = 0, so
x = -1/2 or x = 3.
Then substitute each of these back in, to get a corresponding value of y for each solution.
4x^2 + 2x = 2y - 8, and subtract that from the 1st equation:
2x^2 - 5x = 3. Collecting terms on 1 side, you have:
2x^2 - 5x - 3 = 0, or
(2x + 1)(x - 3) = 0, so
x = -1/2 or x = 3.
Then substitute each of these back in, to get a corresponding value of y for each solution.
2015-05-24 1:08 am
6x^2 - 3x = 2y - 5
2x^2 + x = y - 4.
4x^2 + 2x = 2y - 8
6x^2 - 3x + 5 = 4x^2 + 2x + 8
2X^2 - 5x - 3 = 0
2x^2 - 6x + x - 3 = 0
2x(x - 3) + (x - 3) = 0
(2x + 1)(x - 3) = 0
Solutions:
x = -1/2, y = 4
x = 3, y = 25
2x^2 + x = y - 4.
4x^2 + 2x = 2y - 8
6x^2 - 3x + 5 = 4x^2 + 2x + 8
2X^2 - 5x - 3 = 0
2x^2 - 6x + x - 3 = 0
2x(x - 3) + (x - 3) = 0
(2x + 1)(x - 3) = 0
Solutions:
x = -1/2, y = 4
x = 3, y = 25
2015-05-24 12:31 am
multiply the 2nd equation by -2 and add the two equations together...
(-4x^2-2x=-2y+8)+(6x^2-3x=2y-5) which equals
2x^2-5x=3
2x^2-5x-3=0
2x^2+x-6x-3=0
x(2x+1)-3(2x+1)=0
(x-3)(2x+1)=0 so x=3 and -1/2
using 6x^2-3x=2y-5 with the xs found above we find the corresponding ys
x=3, y=25 (3, 25)
x=-.5, y=4 (-0.5, 4)
(-4x^2-2x=-2y+8)+(6x^2-3x=2y-5) which equals
2x^2-5x=3
2x^2-5x-3=0
2x^2+x-6x-3=0
x(2x+1)-3(2x+1)=0
(x-3)(2x+1)=0 so x=3 and -1/2
using 6x^2-3x=2y-5 with the xs found above we find the corresponding ys
x=3, y=25 (3, 25)
x=-.5, y=4 (-0.5, 4)
收錄日期: 2021-04-21 01:52:37
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