Question 1128867
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<pre>
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.


<U>Answer</U>.  x= -14;  y= -3.
</pre>

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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