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

[juantheron]

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)=

 

[Opalg]

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.