How do I go about finding x and y in:
-1 - y
-------- = -1/3
x - 3y
2006-11-17 08:41:09 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Ooops. I totally typed that wrong it should have been:
-1 - y
-------- = -1/3
x + 4y
Is the answer any value where x and y = 3, except (4,-1)?
2006-11-17 08:54:37 · update #1
( -1 -y) / (x + 4y) = -1/3
=> (1 + y) = (x + 4y)/3
=> 3 + 3y = x + 4y
=> x + y = 3 => x = 3 - y = k =>
x = k
y = 3 - k
it means,for infinite number of k,x = k and y = 3 - k
2006-11-17 09:59:35 · answer #1 · answered by farbod f 2 · 0⤊ 0⤋
Answer to your corrected part
You are right as equation becomes x+y=3, but lower part is 0 when (4,-1) so that is ruled out.
------------
if u r looking fior the exact answer, its not possible in this case. But if you have
1+y
------= -1/3
x-3y
Then x = -3, but Y can not be found.
Basicaly, to solve this equation, you need to provide another equation. An equation with n variables needs minimum n-1 more equations to solve it.
However, if you are looking only for integral solution for your original question, here is your solution
x = 6y+3 (I have shortened your equation)
means give any value for y and you get x. so your solution pairs are
(x,y) = .... (-9,-2) or (-3,-1) or (3,0) or (9,1) or (15,2) and similar such pairs.
2006-11-17 17:00:10 · answer #2 · answered by golgolbaat 3 · 1⤊ 0⤋
x=1
y=2
first you should cross multiply to find the value of x
-1-y
----- = -1/3
x+4y
-3-3y= -x-4y
x= - y+3
then substitute the value for x and cross multiply
-1-y
------ = -1/3
-y+3+4y
-3 - 3y= -1/3
-3(1+y)= - 1/3
1+y=1
y=2
Then substitute y to find x
x = -2 +3
x = 1
2006-11-17 17:18:01 · answer #3 · answered by RDEEMD 2 · 0⤊ 0⤋
the actual answer is...
if x=3,then y is any real number other than -1.... if y=-1 for x=3 the LHS of the expression will become 0/0 form......
2006-11-17 16:53:02 · answer #4 · answered by fool 1 · 0⤊ 0⤋
Clear the denominator by multipling both sides. You then have and equation with just x, y and coefficients. You will get a line.
2006-11-17 16:47:29 · answer #5 · answered by modulo_function 7 · 0⤊ 0⤋
-1 -y =-x/3 + y
x/3 - 1 = 2y
x = 3(2y + 1)
2006-11-17 16:50:06 · answer #6 · answered by sloop_sailor 5 · 0⤊ 0⤋
cross multiply
(-1-y) (3) = (-1) (x - 3y)
-3 - 3y = -x + 3y
x = 3
but +3y - 3y cancels out
therefore y can be any real number
x = 3
y = All real numbers
2006-11-17 16:48:29 · answer #7 · answered by naike_10021980 2 · 0⤊ 1⤋
x = -3.
y can be any number. Equation holds good for any value of y: -infinity < y < infinity
2006-11-17 16:46:17 · answer #8 · answered by ramshi 4 · 0⤊ 0⤋