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From the condition, we have these two equations
2x = x + y -11 (1)
3y + 2 = x+ 7 (2)
Simplify each equation
x - y = -11 (1')
-x + 3y = 5 (2')
Add equations (1') and (2'). You will get
2y = -11 + 5 = -6 ====> y = -6/2 = -3.
Then from eq(1') x = y - 11 = -3 - 11 = -14.
Answer. x= -14; y= -3.
Solved.
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@josgarithmetic correctly noticed that with these values of x and y the dimensions of the "rectangle" are negative,
so, such a "rectangle" does not exist.
It is true.
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