To solve the system of equations given, let’s denote the equations as follows:
From Eq1 + Eq2:
((x+y+z+w) + (x+y-z-w) = 4+2)
(2x + 2y = 6 \implies x + y = 3) ...(Eq5)
From Eq3 + Eq4:
((x-y+z-w) + (x-y-z+w) = 0+0)
(2x -2y =0 \implies x = y) ...(Eq7)
From Eq5 and Eq7:
(x + x =3 \implies x = \frac{3}{2}, y = \frac{3}{2})
From Eq1 - Eq2:
((x+y+z+w) - (x+y-z-w) =4-2)
(2z +2w =2 \implies z +w =1) ...(Eq6)
From Eq3 - Eq4:
((x-y+z-w) - (x-y-z+w)=0)
(2z -2w=0 \implies z =w) ...(Eq8)
From Eq6 and Eq8:
(z + z =1 \implies z = \frac{1}{2}, w = \frac{1}{2})
(x = \frac{3}{2}, y = \frac{3}{2}, z = \frac{1}{2}, w = \frac{1}{2})
Answer:
(x=1.5, y=1.5, z=0.5, w=0.5) or in fractions (\boxed{x=\frac{3}{2}, y=\frac{3}{2}, z=\frac{1}{2}, w=\frac{1}{2}})
If the problem expects integer form (though not possible here), but as per the solution, the values are these fractions.
Simplified numerical answer: (\boxed{1.5,1.5,0.5,0.5}) (but likely fractions are preferred: (\boxed{\frac{3}{2},\frac{3}{2},\frac{1}{2},\frac{1}{2}}))
(\boxed{x=\frac{3}{2}, y=\frac{3}{2}, z=\frac{1}{2}, w=\frac{1}{2}})
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