[tex]{(x + y - 1)}^{2} + {(4x - 6y + 1)}^{2} = 0[/tex]

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вопрос

This equation represents the sum of squares of two expressions, which means that the only way for the sum to be equal to zero is if both expressions are equal to zero.

So we have:

[ x + y - 1 = 0
[ 4x - 6y + 1 = 0 ]

From the first equation, we can solve for x:

[ x = 1 - y ]

Substitute this into the second equation:

[ 4(1 - y) - 6y + 1 = 0
[ 4 - 4y - 6y + 1 = 0
[ 5 - 10y = 0
[ 10y = 5
[ y = \frac{1}{2} ]

Now, substitute y back into the first equation to find x:

[ x = 1 - \frac{1}{2}
[ x = \frac{1}{2} ]

Therefore, the solution to this equation is ( x = \frac{1}{2} ) and ( y = \frac{1}{2} ).