Please solve these equations and find x y z and p

Question

2x=3y+1

2y=3z+1

2z=3p+1

Answer 1

2x=3y+1

2y=3z+1

2z=3p+1

x=(3y=1)/2

y=(3z+1)/2

z=(3p+1)/2

 

using y=(3z+1)/2, replace z by (3p+1)/2

y=[(3)[(3p+1)/2]+(1)]/(2)

y=[(9p+3)/2+1]/2

y=[(9/2)p+(3/2)+1]/2

y=[(9/2)p+(5/2)]/2

y=[(9p+5)/2]/2

y=(9p+5)/4

let p=3

y=(27+5)/4

y=32/4

y=8

2x=3*8+1

2x=25

x=12.5

2y=3z+1

2*8=3z+1

16=3z+1

3z=15

z=5

x=12.5, y=8, z=5, and p=3

25=24+1

16=15+1

10=9+1

Answer 2

At the moment, you have 3 equations in four unknowns, so you'll have to do it by substitutions, considering p to be a constant (or parameter).  Awkward, but you can write y as a function of x, z as a function of y (and thus ultimately of x), and z as a function of p.