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system of equations

jacks

Well-known member
Apr 5, 2012
226
If [tex]a\;,b\;,x\;,y[/tex] satisfy [tex]\begin{Bmatrix} ax+by=4\\\\
ax^2+by^2=-3 \\\\
ax^3+by^3=-3 \\\\
\end{Bmatrix}[/tex]. Then [tex](2x-1)(2y-1)=[/tex]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
If [tex]a\;,b\;,x\;,y[/tex] satisfy [tex]\begin{Bmatrix} ax+by=4\\\\
ax^2+by^2=-3 \\\\
ax^3+by^3=-3 \\\\
\end{Bmatrix}[/tex]. Then [tex](2x-1)(2y-1)=[/tex]
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.