Question

não consigo resolver essaaa ... ajudem, por favor

Answer 1

$(k-3)*x + \frac{2k-5}{4} +4k =0\\\\ 4*\left[(k-3)*x + \frac{2k-5}{4} +4k\right] = 4*0\\\\(k-3)*4x+2k-5+4*4k=0\\\\(k-3)*4x+2k-5+16k=0\\\\(k-3)4x +18k-5=0\\\\. \boxed{x=3}\\\\(k-3)*4*3 +18k -5 =0\\\\12k -36+18k-5=0\\\\30k -41 =0\\\\ k= \frac{41}{30}$