Compute all ordered pairs of integers (x,y) that satisfy xy + 4x - 13 = x^2.
Thanx!
I need some help with this homework question.?
2008-01-08 5:01 am
回答 (12)
2008-01-08 5:46 am
✔ 最佳答案
Robby R and DW each *almost* got it right.Notice that xy = x² -4x + 13, so either x=0 or
y = x - 4 + 13/x.
Now, for this to be an integer, we must have 13/x be an integer, so x = ±1, ±13. Thus the solutions are
(-13, -18), (-1, -18), (1, 10), and (13, 10).
(x=0 produces no solutions, as you can check.)
§
2008-01-08 1:12 pm
xy=x^2-4x+13
If x is not 0, y=x-4+13/x
The only answers with integers for y are x=-13 and x=13, otherwise 13/x is not going to be an integer.
so,
(-13,-18)
(13,10)
Then test the x=0 case:
0*y+4(0)-13=0^2, or
-13=0. This is clearly not true, so x=0 is not a solution.
If x is not 0, y=x-4+13/x
The only answers with integers for y are x=-13 and x=13, otherwise 13/x is not going to be an integer.
so,
(-13,-18)
(13,10)
Then test the x=0 case:
0*y+4(0)-13=0^2, or
-13=0. This is clearly not true, so x=0 is not a solution.
2008-01-12 1:01 pm
xy + 4x - 13 = x²
x² - 4x -xy + 13 = 0
x² - (4+y)x + 13 = 0
(x-13)(x-1) = x² - 14x + 13
x² - 14x + 13 = x² - (4-y)x + 13
â 4+y = 14
â y = 10
Answers are (1,10) and (13,10)
x² - 4x -xy + 13 = 0
x² - (4+y)x + 13 = 0
(x-13)(x-1) = x² - 14x + 13
x² - 14x + 13 = x² - (4-y)x + 13
â 4+y = 14
â y = 10
Answers are (1,10) and (13,10)
2008-01-08 1:34 pm
xy + 4x - 13 = x²
x² - 4x -xy + 13 = 0
x² - (4+y)x + 13 = 0
Since 13 is prime and x must be an integer, the only possible roots are x=13 and x=1.
(x-13)(x-1) = x² - 14x + 13
x² - 14x + 13 = x² - (4-y)x + 13   Â
â   4+y = 14
â   y = 10
(1,10)
(13,10)
x² - 4x -xy + 13 = 0
x² - (4+y)x + 13 = 0
Since 13 is prime and x must be an integer, the only possible roots are x=13 and x=1.
(x-13)(x-1) = x² - 14x + 13
x² - 14x + 13 = x² - (4-y)x + 13   Â
â   4+y = 14
â   y = 10
(1,10)
(13,10)
2008-01-08 1:22 pm
Two of the answers are (1,10) and (3,4). I'll keep going for a while.
What I have worked out so far (I think) is that the x value has to be odd and the y value has to be even.
The function is a hyperbola, with a vertical asymptote at x = 0.
What I have worked out so far (I think) is that the x value has to be odd and the y value has to be even.
The function is a hyperbola, with a vertical asymptote at x = 0.
2008-01-08 1:04 pm
tough prob 4 me
2008-01-08 1:17 pm
well it depends wat you want to solve for, like if you want to solve for y first, move everything without the y to one side, and divde the xy by x. then do the same thing for x.
2008-01-08 1:07 pm
chill a bit, amigo just put the value of x as 0,1,2,3,------and get the value of correspondingly ie there will be infinite solutions
2008-01-08 1:06 pm
Um im no genuis im in algebra 2 but u do the square root on both sides then simplify more i hope i made sence
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