System of Equations: (2x-1)(2y-1)

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SUMMARY

The discussion revolves around solving a system of equations involving variables \(a\), \(b\), \(x\), and \(y\) defined by the equations \(ax + by = 4\), \(ax^2 + by^2 = -3\), and \(ax^3 + by^3 = -3\). The values for \(a\) and \(b\) are derived as \(a = \frac{4y + 3}{x(y - x)}\) and \(b = \frac{-4x - 3}{y(y - x)}\). Substituting these into the third equation leads to the expression \(4xy + 3x + 3y - 3 = 0\). The conclusion drawn is that \((2x - 1)(2y - 1) = 4 - 5(x + y)\), which is deemed unsatisfactory compared to the alternative expression \(\left(2x + \frac{3}{2}\right)\left(2y + \frac{3}{2}\right)\).

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If a\;,b\;,x\;,y satisfy \begin{Bmatrix} ax+by=4\\\\ <br /> ax^2+by^2=-3 \\\\ <br /> ax^3+by^3=-3 \\\\ <br /> \end{Bmatrix}. Then (2x-1)(2y-1)=
 
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jacks said:
If a\;,b\;,x\;,y satisfy \begin{Bmatrix} ax+by=4\\\\ <br /> ax^2+by^2=-3 \\\\ <br /> ax^3+by^3=-3 \\\\ <br /> \end{Bmatrix}. Then (2x-1)(2y-1)=
Solve the first two equations for $a$ and $b$. You should get $a = \dfrac{4y+3}{x(y-x)}$, $b = \dfrac{-4x-3}{y(y-x)}.$ Now substitute those values of $a$ and $b$ into the third equation. That should lead to $4xy+3x+3y-3=0.$

At that point I run into trouble. The best that you can deduce about $(2x-1)(2y-1)$ from that last equation is that $(2x-1)(2y-1) = 4-5(x+y)$, which does not seem like a satisfactory answer. If instead the question had asked for $\bigl(2x+\frac32\bigr)\bigl(2y+\frac32\bigr)$ then you could have re-written $4xy+3x+3y-3$ as $\bigl(2x+\frac32\bigr)\bigl(2y+\frac32\bigr) -\frac{21}4$ to give the answer $21/4$, which somehow seems to be more like the sort of conclusion that the question calls for.
 

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