x+y−2z = 0

3x+y = 1

5x+3y+7z = 2

(2) y = 1 – 3x

–>(1) x + (1 – 3x) – 2z = 0

-2x – 2z + 1 = 0

2x + 2z = 1

x + z = 1/2

x = 1/2 – z (1b)

–>(3) 5(1/2 – z) + 3{1 – 3(1/2 – z)} + 7z = 2

5/2 – 5z + 3 – 9(1/2 – z) + 7z = 2

5/2 – 5z + 3 – 9/2 + 9z + 7z = 2

-2 + 3 + 11z = 2

1 + 11z = 2

11z = 1

z = 1/11

–>(1) x + y – 2(1/11) = 0

x + (1 – 3x) – 2(1/11) = 0

x + 1 – 3x -2/11 = 0

-2x + 9/11 = 0

x = 9/22

–>(2) y = 1 – 3(9/22)

y = 1 – 27/22

y = -5/22

x = 9/22

y = -5/22

z = 1/11

Check :

x + y – 2z = 0

9/22 + (-5/22) – 2(1/11) = 0

9/22 – 5/22 – 2(2/22) = 0

9/22 – 5/22 – 4/22 = 0

0 = 0

3(9/22) + (-5/22) = 1

27/22 – 5/22 = 1

22/22 = 1

1 = 1

5(9/22) + 3(-5/22) + 7(1/11) = 2

45/22 – 15/22 + 14/22 = 2

(45 – 15 + 14)/22 = 2

44/22 = 2

2 = 2

needless to say y and z can not the two be even because of fact this makes LHS even and RHS strange. Can y, z be equivalent? if so then x*y^2 = 2y^2 + a million —-> (x – 2)*(y^2) = a million and clearly x = 3, y = a million is the only answer with integers. to any extent further i visit anticipate that y is the bigger of y, z as reversed suggestions are incredibly the same. Can z = a million be a answer without y = a million? xy = y^2 + 2 —-> y^2 – xy + 2 = 0 —-> y = [x + sqrt(x^2 – 8)]/2 x = 3 leads to y = 2 so x = 3, y = 2, z = a million is a answer with all 3 diverse. Can z = 2? 2xy = y^2 + 5 —-> y^2 – 2xy + 5 = 0 —-> y = [2x + sqrt(4x^2 – 20)]/2 = x + sqrt(x^2 – 5) This has answer x = 3, y = 5, z = 2. For sqrt(x^2 – 5) to be an integer we want x^2 – 5 = ok^2 yet 2^2 and 3^2 are the only squares differing by ability of five so this leads to no different suggestions. I easily have discovered, yet won’t placed the evidence here, that z (the smaller of y, z undergo in concepts) can not be 3 or 4. This makes me suspect that there are no longer the different suggestions. EDIT. That final line is inaccurate! i’ve got only discovered that x = 3, y = 13, z = 5 is a answer. in spite of the undeniable fact that, this is calling as though x would desire to be 3.

x+y−2z = 0

3x+y = 1

5x+3y+7z = 2

three equations and three unknowns! WOOHOO!

x=-y+2z

3(-y+2z)+y = 1

-3y+6z+y=1

-2y+6z=1

6z=1+2y

z=(1+2y)/6

5(-y+2((1+2y)/6))+3((1+2y)/6)+7((1+2y)/6) = 2

-5y-10((1+2y)/6)+3((1+2y)/6)+7((1+2y)/6)=2

-5y=2

y=-2/5

z=(1+2y)/6 = (1-4/5)/6 = 1/30

x=-y+2z=2/5+2/30=14/30=7/15