Answer:
[tex] x = 0 [/tex] or [tex] x = -\dfrac{11}{3} [/tex]
Step-by-step explanation:
I assume the denominators are x + 5 and x + 4 although that is not what you wrote.
[tex] \dfrac{4x}{x+5} = \dfrac{x}{x+4} [/tex]
[tex] (x + 4)(x + 5)\times \dfrac{4x}{x+5} = (x + 4)(x + 5) \times \dfrac{x}{x+4} [/tex]
[tex] (x + 4)(4x) = (x + 5)(x) [/tex]
[tex] 4x^2 + 16x = x^2 + 5x [/tex]
[tex] 3x^2 + 11x = 0 [/tex]
[tex] x(3x + 11) = 0 [/tex]
[tex] x = 0 [/tex] or [tex] 3x + 11 = 0 [/tex]
[tex] x = 0 [/tex] or [tex] 3x = -11 [/tex]
[tex] x = 0 [/tex] or [tex] x = -\dfrac{11}{3} [/tex]
