OR Tools Constraint that requires equility of variables [ortools] - or-tools

Question

I want create a linear solver model using OR-Tools. I have two numVars and one of my constraints is the equility of these two variables. But I could not find a way to set a constraint that takes a variable. I can do this with gurobi library. Below are the codes for both libraries. Is there a way to pass variables to both sides of a constraint such as xa==xb?
Thanks in advance.
Gurobi Code:
from gurobipy import Model, GRB
m = Model('rafinery')
xa = m.addVar(vtype=GRB.CONTINUOUS)
xb = m.addVar(vtype=GRB.CONTINUOUS)
# production constraint
m.addConstr(xa*400 + xb*300 >= 25000)
m.addConstr(xa*300 + xb*400 >= 27000)
m.addConstr(xa*200 + xb*500 >= 30000)
m.addConstr(xa>=0)
m.addConstr(xb>=0)
m.addConstr(xa==xb)
m.setObjective(xa * 20000 + xb * 25000, GRB.MINIMIZE)
m.update()
m.setObjective(xa + xb, GRB.MINIMIZE)
m.update()
m.optimize()
for v in m.getVars():
print(v.varName, v.x)
print("Cost:", m.objVal)
OR-Tools Code:
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveSimpleSystem', pywraplp.Solver.GLOP_LINEAR_PROGRAMMING)
# Create Variables
xa = solver.NumVar(0, solver.infinity(),'xa')
xb = solver.NumVar(0, solver.infinity(),'xb')
# Create Constraints
# xa*400 + xb*300 >= 25000
high = solver.Constraint(25000, solver.infinity())
high.SetCoefficient(xa, 400)
high.SetCoefficient(xb, 300)
# xa*300 + xb*400 >= 27000
middle = solver.Constraint(27000, solver.infinity())
middle.SetCoefficient(xa,300)
middle.SetCoefficient(xb,400)
# xa*200 + xb*500 >= 30000
high = solver.Constraint(30000, solver.infinity())
high.SetCoefficient(xa, 200)
high.SetCoefficient(xb, 500)
# another constraint that factories work for same days
# ******************** get error here ****************
noidle = solver.Constraint(xa,xa)
noidle.SetCoefficient(xb,1)
# obj = minimize production cost which is
# xa * 20000 + xb * 25000
obj = solver.Objective()
obj.SetCoefficient(xa, 20000)
obj.SetCoefficient(xb, 25000)
obj.SetMinimization()
solver.Solve()
print("Rafinery Solution:")
print("Rafinery A work days:", xa.solution_value())
print("Rafinery B work days:", xb.solution_value())
print("Cost : ", solver.Objective().Value())

Answer 1

from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveSimpleSystem', pywraplp.Solver.GLOP_LINEAR_PROGRAMMING)
# Create Variables
xa = solver.NumVar(0, solver.infinity(),'xa')
xb = solver.NumVar(0, solver.infinity(),'xb')
# production constraint
solver.Add(xa*400 + xb*300 >= 25000)
solver.Add(xa*300 + xb*400 >= 27000)
solver.Add(xa*200 + xb*500 >= 30000)
solver.Add(xa>=0)
solver.Add(xb>=0)
solver.Add(xa==xb)
solver.Minimize(xa * 20000 + xb * 25000)
solver.Solve()
print("Rafinery Solution:")
print("Rafinery A work days:", xa.solution_value())
print("Rafinery B work days:", xb.solution_value())
print("Cost : ", solver.Objective().Value())