check the picture below.
in the rectangle, all sides meet at right-angles, that makes the two pairs of sides a pair of two equal sides. So the left and right are equal and top and bottom are also equal, as in the picture below.
[tex]\bf \begin{cases}
12y+10=10x\\
6x+3=9y
\end{cases}\\\\
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12y+10=10x\implies 12y=10x-10\implies y=\cfrac{10x-10}{12}
\\\\\\
y=\cfrac{10x}{12}-\cfrac{10}{12}\implies y=\cfrac{5x}{6}-\cfrac{5}{6}\implies \boxed{y=\cfrac{5x-5}{6}}\\\\
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\textit{now, substituting in the \underline{second function}}[/tex]
[tex]\bf 6x+3=9\left( \boxed{\cfrac{5x-5}{6}} \right) 6x+3=3\cdot \cfrac{5x-5}{2}
\\\\\\
6x+3=\cfrac{15x-15}{2}\implies 12x+6=15x-15\implies 21=3x
\\\\\\
\cfrac{21}{3}=x\implies \boxed{7=x}\\\\
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\textit{now, we know that }y=\cfrac{5x-5}{6}\implies y=\cfrac{5(7)-5}{6}
\\\\\\
y=\cfrac{35-5}{6}\implies y=\cfrac{30}{6}\implies \boxed{y=5}[/tex]
