Check the picture below.
[tex]\bf \stackrel{\measuredangle DAC}{4y+2x}~~=~~\stackrel{\measuredangle BCA}{9y-x}\implies 2x=5y-x\implies 3x=5y\implies \boxed{x=\cfrac{5y}{3}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\measuredangle DAB}{[(4y+2x)+35]}~~+~~\stackrel{\measuredangle ADC}{5x+4}~~=~~180\implies 4y+2x+5x+39 = 180 \\\\\\ 4y+7x+39=180\implies 4y+7x=141\implies \stackrel{\textit{substituting "x"}}{4y+7\left( \boxed{\cfrac{5y}{3}} \right)} = 141[/tex]
[tex]\bf 4y+\cfrac{35y}{3}=141\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3\left( 4y+\cfrac{35y}{3} \right)=3(141)}\implies 12y+35y=423 \\\\\\ 47y=423\implies y=\cfrac{423}{47}\implies \blacktriangleright y = 9 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{5y}{3}}\implies x = \cfrac{5(9)}{3}\implies \blacktriangleright x = 15 \blacktriangleleft[/tex]
