Given that the figure is an isosceles trapezoid,
Isosceles trapezoid is a trapezoid whose two opposite sides and two adjacent angles are congruent.
Therefore,
[tex]\begin{gathered} 14x-25=19y+10 \\ 14x-19y=10+25 \\ 14x-19y=35\ldots\ldots\ldots.1 \end{gathered}[/tex]
Also,
[tex]\begin{gathered} \angle U+\angle T=180 \\ 3x+1+19y_{}+10=180 \\ 3x+19y+11=180 \\ 3x+19y=180-11 \\ 3x_{}+19y=169\ldots\ldots\ldots\ldots.2 \end{gathered}[/tex]
Combining the two equations together and solving for x and y
[tex]\begin{gathered} 14x-19y=35\ldots\ldots\ldots.1 \\ 3x_{}+19y=169\ldots\ldots\ldots\ldots.2 \end{gathered}[/tex]
Isolate x from equation 1
[tex]\begin{gathered} 14x=35_{}+19y \\ \therefore x=\frac{35+19y}{14} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Substitute\: }x=\frac{35+19y}{14}\text{ into equation 2} \\ 3\times\frac{35+19y}{14}+19y=169 \\ \frac{105+323y}{14}=169 \end{gathered}[/tex]
Solve for y
[tex]\begin{gathered} 105+323y=169\times14 \\ 323y=169\times14-105 \\ 323y=2261 \\ y=\frac{2261}{323}=7 \\ \therefore y=7 \end{gathered}[/tex][tex]\begin{gathered} \mathrm{For\: }x=\frac{35+19y}{14} \\ \mathrm{Substitute\: }y=7 \\ x=\frac{35+19\times7}{14}=\frac{35+133}{14}=\frac{168}{14}=12 \\ \therefore x=12 \end{gathered}[/tex]
Hence,
[tex]x=12,y=7[/tex]
