From the given figure,
[tex]\Delta PQR\text{ is similar to }\Delta STU[/tex]
When two triangles are similar then ratio of their corresponding sides are similar.
[tex]\frac{PQ}{ST}=\frac{QR}{TU}=\frac{PR}{SU}[/tex]
Substituting the values in the above equation,
[tex]\frac{8}{16}=\frac{10}{x\text{ + 3}}=\frac{5y-1}{28}[/tex]
Calculating the value of x ,
[tex]\begin{gathered} \frac{8}{16}=\text{ }\frac{10}{x+3} \\ 8(x+3)=10\times16 \\ 8x+24=160 \end{gathered}[/tex]
Rearranging the like terms o both the sides ,
[tex]\begin{gathered} 8x\text{ = 160 - 24} \\ 8x\text{ = 136} \\ x\text{ = }\frac{136}{8} \\ x\text{ = 17} \end{gathered}[/tex]
Calculating the value of y,
[tex]\begin{gathered} \frac{8}{16}\text{ = }\frac{5y-1}{28} \\ 8\text{ }\times\text{ 28 = }16\times\text{ (5y-1)} \\ 224\text{ = 80y - 16} \\ \end{gathered}[/tex]
Rearranging the like terms on both the sides ,
[tex]\begin{gathered} 224\text{ + 16 = 80y} \\ 80y\text{ = 240} \\ y\text{ = }\frac{240}{80} \\ y\text{ = 3} \end{gathered}[/tex]
Thus the required values of x and y are ,
[tex]\begin{gathered} x\text{ = 17} \\ y\text{ = 3} \end{gathered}[/tex]
