Check the picture below.
let's change the decimal amounts to fractions, so for 3.5 let's use 7/2 and for 1.5 let's use 3/2, let's use the proportion on the left side and then the next one after
[tex]\cfrac{3.5}{2x+7}=\cfrac{x-1.5}{8}\implies \cfrac{~~ \frac{7}{2}~~}{2x+7}=\cfrac{x-\frac{3}{2}}{8}\implies \cfrac{~~ \frac{7}{2}~~}{\frac{2x+7}{1}}=\cfrac{\frac{2x-3}{2}}{\frac{8}{1}} \\\\\\ \cfrac{7}{2}\cdot \cfrac{1}{2x+7}=\cfrac{2x-3}{2}\cdot \cfrac{1}{8}\implies \cfrac{7}{4x+14}=\cfrac{2x-3}{16} \\\\\\ 112=8x^2-12x+28x-42\implies 0=8x^2+16x-154 \\\\\\ 0=2(4x^2+8x-77)\implies 0=4x^2+8x-77\implies 0=(2x-7)(2x+11) \\\\\\ 7=2x\implies \boxed{\cfrac{7}{2}=x}[/tex]
now, let's notice that we didn't use the 2x+11, since that gives us a negative "x" and "x" cannot be a negative value.
[tex]\cfrac{3.5}{x-1.5}=\cfrac{y}{x+6}\implies \cfrac{~~\frac{7}{2}~~}{\frac{7}{2}-\frac{3}{2}}=\cfrac{y}{\frac{7}{2}+6}\implies \cfrac{~~ \frac{7}{2}~~}{2}=\cfrac{y}{~~\frac{19}{2}~~} \\\\\\ \cfrac{7}{4}=\cfrac{2y}{19}\implies 133=8y\implies \cfrac{133}{8}=y\implies 16\frac{5}{8}=y=BC[/tex]
