the numbers: x, y
\(x=3y-11 \\
x+y=24 \\ \\
\hbox{substitute 3y-11 for x in the second equation:} \\
3y-11+y=24 \\
3y+y=24+11 \\
4y=35 \\
y=\frac{35}{4}=8 \frac{3}{4} \\ \\
x=3 \cdot \frac{35}{4}-11 =\frac{105}{4}-\frac{44}{4}=\frac{61}{4}=15 \frac{1}{4}\)