if x-y=8,y=3z and z=2 what is the value of x?

Answer 1

z = 2
y = 3*2 = 6
x - 6 = 8
x = 14

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Answer 2

needless to say y and z can't the two be even because of fact this makes LHS even and RHS odd. 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. from now on i will think of that y is the bigger of y, z as reversed ideas are fairly the comparable. 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 finally leads to y = 2 so x = 3, y = 2, z = a million is a answer with all 3 distinctive. 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 with the aid of five so this finally leads to no different ideas. I even have got here across, yet won't positioned the evidence right here, that z (the smaller of y, z remember) won't be able to be 3 or 4. This makes me suspect that there are no longer the different ideas. EDIT. That final line is inaccurate! i've got purely got here across that x = 3, y = 13, z = 5 is a answer. whether, it fairly is calling as though x would desire to be 3.

Answer 3

z = 2
y = 3z = 3(2) = 6
x-y = 8
x - 6 = 8
x = 8+6
x = 14

Answer 4

14

Answer 5

14

Answer 6

substitute 2 for z :

y = 3 (2)
y = 6

then substitute 6 for y :

x - 6 = 8
+6 = +6

x = 14

Answer 7

z = 2

y = 3z
y = 3(2)
y = 6

x - 6 = 8
x = 8 + 6
x = 14

Answer 8

X-Y=8
y=3z
z=2

all you have to do is fill in!

y=3(2)
y=6

x-6=8
x=8+6
x=14

Answer 9

y=3(2)=6
x-6=8
x=14

Answer 10

y=3(2)=6
x-6=8
x=14