Given:

\(\displaystyle{m}∠{A}{B}{D}={7}{x}−{5}\)

\(m∠CBD=4x+1\)

BD bisects \(\displaystyle∠{A}{B}{C}\)

x=?

When BD bisects \(\displaystyle∠{A}{B}{C}\) then \(\displaystyle∠{A}{B}{D}=∠{D}{B}{C}\)

7x−5=4x+1

Subtract 4x both the sides:

7x−5−4x=4x+1

3x−5=1

Add 5 both the sides:

3x−5+5=1+5

3x=6

Divide 3 by both the sides:

3x3=63

x=2

Hence value of x ix 2