-8x+5y = 18 -> -48x + 30y = 108 (1)
7x + 6y = 5 -> 35x + 30y = 25 (2)
(1) - (2) = -48x - 35 x = 108 -25 -> -83x = 83 -> x=-1
-8*-1 + 5y = 18 -> y=2
2007-03-11 08:15:50
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answer #1
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answered by hbj 2
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- 8x + 5y = 18- - - - - -Equation 1
7x + 6y = 5- - - - - - -- Equation 2
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Multiply equation 1 by - 6 and equation 2 by 5
- 8x + 5y = 18
- 6(- 8) + (- 6)(5y) = - 6(18)
48 + (- 30y) = - 108
48 - 30y = - 108
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7x + 6y = 5
5(7x) + 5(6y) = 5(5)
35x + 30y = 25
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48 x - 30y = - 108
35x + 30y= 25
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83x = - 83
83x / 83 = - 83 / 83
x = - 83 / 83
x = - 1
Insert the x value into equaion 1
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- 8x + 5y = 18
- 8( - 1) + 5y = 18
8 + 5y = 18
8 + 5y - 8 = 18 - 8
5y = 10
5y / 5 = 10/ 5
y = 10 / 5
y = 2
Insert the y value into euation 1
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Check for equation 1
- 8x + 5y = 18
- 8(- 1) + 5(2) = 18
8 + 10 = 18
18 = 18
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Check for equation 2
7x + 6y = 5
7(- 1) + 6(2) = 5
- 7 + 12 = 5
5 = 5
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The solution set is { -1, 2 }
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2007-03-11 16:08:26
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answer #2
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answered by SAMUEL D 7
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-8x + 5y = 18
7x + 6y = 5
Multiply the top by y and the bottom by 8:
-56 x + 35 y = 126
56 x + 48y = 40
Now add to isolate y:
83 y = 156 or y = 166/83 = 2
Now that you have y, put it in the bottom equation and solve for x:
7x + 6(2) = 5
7x = 5 - 12 = -7 and x = -1
2007-03-11 15:19:29
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answer #3
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answered by kellenraid 6
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x = -1
y = 2
2007-03-11 15:14:03
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answer #4
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answered by physicist 4
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x= -1, y=2
2007-03-11 15:13:07
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answer #5
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answered by Timothy G 2
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One quick way is to multiply the first equation by the factor of x in the second, then multiply the second by the factor of x in the first:
multiply the first by 7:
-56x +35y = 126
multiply the second by -8:
-56x -48y = -40
then change the sign in one of them (let us say the second one) and "add them up":
-56x +35y = 126
+56x +48y = +40
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0x +83y = 166
then y = 2
Go back and substitute y=2 in any of the equations (better: d both to make sure you get the same value for x)
2007-03-11 15:17:31
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answer #6
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answered by Raymond 7
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That is immposible to answer because it has two unknown variables and not enough to rule on out
2007-03-11 15:16:17
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answer #7
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answered by Anonymous
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