Solution: Find the value of k such that the system has no solutions: 2x+ky equals 6 and 4x+10y equals 12

Code:

import sympy

x, y, k = sympy.symbols('x y k')
eq1 = 2*x + k*y - 6
eq2 = 4*x + 10*y - 12

# A system of linear equations ax + by = c and dx + ey = f has no solutions if:
# a/d = b/e != c/f
# Here: 2/4 = k/10 != 6/12
# 2/4 = 1/2
# 6/12 = 1/2
# If 2/4 = k/10 = 6/12, the lines are identical (infinitely many solutions).
# If 2/4 = k/10 != 6/12, there are no solutions.

# Let's check the ratios:
ratio_x = sympy.Rational(2, 4)
ratio_const = sympy.Rational(6, 12)

print(f"{ratio_x=}")
print(f"{ratio_const=}")

# Solve for k such that 2/4 = k/10
k_val = sympy.solve(sympy.Rational(2, 4) - k/sympy.Integer(10), k)[0]
print(f"{k_val=}")

Output:

ratio_x=1/2
ratio_const=1/2
k_val=5