Statement Simplification

Variable definition :

• Variables
  x∈C
  y∈C
  
• Constants
  p∈N
  

Simplification :

2x+(p+2)y=2∧(p-1)x+2y=1

// Connecting 2x+(p+2)y=2 and (p-1)x+2y=1 using shared item y

• At first, I express y from equation (p-1)x+2y=1
  Transport items not containing y to the right in equation (p-1)x+2y=1
  Divison of equation 2y=1+(-p+1)x with 2
  Equation expression result is y=1+(-p+1)x2
  
• Install to equation 2x+(p+2)y=2
  
• Connected equation is 2x+(p+2)1+(-p+1)x2=2
  

2x+(p+2)1+(-p+1)x2=2∧(p-1)x+2y=1
// There is (p+2)(1+(-p+1)x)2 item in expression 2x+(p+2)(1+(-p+1)x)2 , so i put it in from of the expression.Result is
4x+(p+2)(1+(-p+1)x)2=2∧(p-1)x+2y=1
// Decomposition of summation 1+(-p+1)x in item (p+2)(1+(-p+1)x) .Result is 4x+p1+p(-p+1)x+2+2(-p+1)x Theory
4x+p1+p(-p+1)x+2+2(-p+1)x2=2∧(p-1)x+2y=1
// Put of item x from expression 4x+p+p(-p+1)x+2+2(-p+1)x
x(p(-p+1)+4)+p+2+2(-p+1)x2=2∧(p-1)x+2y=1
// Put of item x from expression x(p(-p+1)+4)+p+2+2(-p+1)x
x(2(-p+1)+p(-p+1)+4)+p+22=2∧(p-1)x+2y=1
// Decomposition of summation -p+1 in item 2(-p+1) .Result is p(-p+1)+4-2p+2 Theory
x(p(-p+1)+4-2p+2)+p+22=2∧(p-1)x+2y=1
// Multiplication of numbers 2 * -1 = -2
// Count of numbers 4 + 2 = 6
x(p(-p+1)+6-2p)+p+22=2∧(p-1)x+2y=1
// Decomposition of summation -p+1 in item p(-p+1) .Result is 6-2p-pp+p1 Theory
x(6-2p-pp+p1)+p+22=2∧(p-1)x+2y=1
// Put of item p from multiplication -pp
x(6-2p-p2+p1)+p+22=2∧(p-1)x+2y=1
// Put of item p from expression 6-2p-p2+p
x(6+p(1-2)-p2)+p+22=2∧(p-1)x+2y=1
// Count of numbers 1 + -2 = -1
x(6-p-p2)+p+22=2∧(p-1)x+2y=1
// Decomposition of quadratic -p2-p+6 , looking for roots for p
// Diskriminant is 25
// Solutions are 2 and -3
// Roots are p-2 and p+3
-x((p-2)(p+3))+p+22=2∧(p-1)x+2y=1
x(-p+2)(p+3)+p+22=2∧(p-1)x+2y=1
// I express x from equation x(-p+2)(p+3)+p+22=2

x=4-p-2(2-p)(p+3)∧(p-1)x+2y=1
// Count of numbers 4 + -2 = 2
(x=2-p(2-p)(p+3)∧2-p≠0∨2-p=0∧4-p-2=0)∧(p-1)x+2y=1
// Count of numbers 4 + -2 = 2
(x=2-p(2-p)(p+3)∧2-p≠0∨2-p=0∧2-p=0)∧(p-1)x+2y=1
// Canceling out 2-p in 2-p(2-p)(p+3)
(x=1(p+3)∧2-p≠0∨2-p=0∧2-p=0)∧(p-1)x+2y=1
// Decomposition of item x=1(p+3)∧2-p≠0∨2-p=0∧2-p=0 in statement (x=1(p+3)∧2-p≠0∨2-p=0∧2-p=0)∧(p-1)x+2y=1 .
(p-1)x+2y=1∧x=1(p+3)∧2-p≠0∨(p-1)x+2y=1∧2-p=0∧2-p=0
// Transport items not containing p to the right in equation 2-p≠0
// Multiplication of p≠-2 with -1
(p-1)x+2y=1∧x=1(p+3)∧p≠2∨(p-1)x+2y=1∧2-p=0∧2-p=0
// Transport items not containing p to the right in equation 2-p=0
// Multiplication of p=-2 with -1
// Transport items not containing p to the right in equation 2-p=0
// Multiplication of p=-2 with -1
(p-1)x+2y=1∧x=1(p+3)∧p≠2∨(p-1)x+2y=1∧p=2∧p=2
(p-1)x+2y=1∧x=1(p+3)∧p≠2∨(p-1)x+2y=1∧p=2
// Connecting (p-1)x+2y=1 and x=1(p+3) using shared item x

• Use of equation x=1(p+3)
  Install to equation (p-1)x+2y=1
  
• Connected equation is p-1p+3+2y=1
  

// Connecting (p-1)x+2y=1 and p=2 using shared item p

• Use of equation p=2
  Install to equation (p-1)x+2y=1
  
• Connected equation is (2-1)x+2y=1
  

p-1p+3+2y=1∧x=1(p+3)∧p≠2∨(2-1)x+2y=1∧p=2
// Count of numbers 2 + -1 = 1
p-1p+3+2y=1∧x=1(p+3)∧p≠2∨1x+2y=1∧p=2
// There is p-1p+3 item in expression p-1p+3+2y , so i put it in from of the expression.Result is
p-1+(p+3)2yp+3=1∧x=1(p+3)∧p≠2∨x+2y=1∧p=2
// I express y from equation (p+3)2y+p-1p+3=1

y=1(p+3)-p+12(p+3)∧x=1(p+3)∧p≠2∨x+2y=1∧p=2
// Count of numbers 1 + 3 = 4
x=1(p+3)∧p≠2∧y=-p+4+p2(p+3)∧p+3≠0∨x+2y=1∧p=2
// Put of item p from expression -p+4+p
x=1(p+3)∧p≠2∧y=p(1-1)+42(p+3)∧p+3≠0∨x+2y=1∧p=2
// Count of numbers 1 + -1 = 0
x=1(p+3)∧p≠2∧y=p0+42(p+3)∧p+3≠0∨x+2y=1∧p=2
// Count of numbers 0 + 4 = 4
x=1(p+3)∧p≠2∧y=42(p+3)∧p+3≠0∨x+2y=1∧p=2
// Put of item 2 from multiplication 22(p+3)2
2(2-1)p+3
x=1(p+3)∧p≠2∧y=2(2-1)p+3∧p+3≠0∨x+2y=1∧p=2
// Count of numbers 2 + -1 = 1
x=1(p+3)∧p≠2∧y=21p+3∧p+3≠0∨x+2y=1∧p=2
// Transport items not containing p to the right in equation p+3≠0
x=1(p+3)∧p≠2∧y=2p+3∧p≠-3∨x+2y=1∧p=2

Result: x=1(p+3)∧p≠2∧y=2p+3∧p≠-3∨x+2y=1∧p=2