Find all the ordered pairs of real numbers that satisfy the following
x+yz=6
y+xz=6
z+ xy=6
2006-10-01 17:19:38 · 9 answers · asked by Antonio G 1 in Science & Mathematics ➔ Mathematics
An ordered pair of real numbers contains two numbers. You specified three equations in three unknowns. This would require an ordered triplet (2,2,2) (1,1,5) (1.5.1) or (5,1,1).
2006-10-01 17:29:34 · answer #1 · answered by Richard 7 · 62⤊ 0⤋
The possibilities are
x = -3, 1, 1, 2, 5
y = -3, 1, 5, 2, 1
z = -3, 5, 1, 2, 1
I went to quickmath.com, click on Solve under Equations, then clicked Advanced and typed in your problem and i made sure x,y,z was on the right hand side, i clicked solve and thats what it gave me.
2006-10-01 17:26:10 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
Sherman81 is giving solutions in integers only (Diophantine equations). Your question wants solutions in real numbers.
2006-10-01 17:33:56 · answer #3 · answered by williamh772 5 · 0⤊ 0⤋
you dont mean pairs....
you mean more like 3 numbers: (x,y,z)
try x=2,y=2,z=2
2006-10-01 17:31:16 · answer #4 · answered by Anonymous · 0⤊ 0⤋
maybe (2,2,2)
substitute 2's in for all x, y, and z and it makes the statement true...don't know if there are anymore...guess and check
2006-10-01 17:24:12 · answer #5 · answered by big_j_gizzy 4 · 0⤊ 0⤋
x+yz=y+xz
x-y = z(x-y)
z=1 if x<>y
y+xz = z+xy
y-z = x(y-z)
x=1 if z<>y
y=5
x,y, z,x, y,z
1,1, 5,1 1,5
1,5, 1,1 5,1
5,1, 1,5 1,1
2,2, 2,2 2,2
2006-10-01 17:47:16 · answer #6 · answered by Helmut 7 · 0⤊ 0⤋
My quick guess was also 2,2,2
2006-10-01 17:32:12 · answer #7 · answered by george_niles 2 · 0⤊ 0⤋
that's too easy... x=2.... y=2........ z=2....
2006-10-01 18:19:00 · answer #8 · answered by the chosen one 1 · 0⤊ 0⤋
2,2,2
2006-10-01 17:46:39 · answer #9 · answered by ag_iitkgp 7 · 0⤊ 0⤋