How to Solve for Unknown Variables with Given Equations?

[anemone]

Here is this week's POTW:

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Evaluate $2014(x+y-xy)-100(a+b)$ if

$ax+by=7\\ax^2+by^2=49\\ax^3+by^3=133\\ax^4+by^4=406$

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[anemone]

Congratulations to the following members for their correct solutions!(Cool)

1. castor28
2. kaliprasad
3. Olinguito
4. Opalg

Solution from Opalg:

Spoiler

$$(1) \qquad ax+by=7\\ (2)\qquad ax^2+by^2=49\\(3)\qquad ax^3+by^3=133\\(4)\qquad ax^4+by^4=406$$ Subtract $y$ times (1) from (2), $y^2$ times (1) from (3), and $y^3$ times (1) from (4): $$(5)\qquad ax(x-y) = 7(7-y), \\ (6)\qquad ax(x^2-y^2) = 7(19-y^2),\\ (7)\qquad ax(x^3-y^3) = 7(58-y^3).$$ Now divide (6) by (5) to get $x+y = \dfrac{19-y^2}{7-y}.$ Then $(x+y)(7-y) = 19-y^2$, so that $$(8)\qquad 7(x+y) - xy = 19.$$ In the same way, divide (7) by (5) to get $(x^2 + xy + y^2)(7-y) = 58 - y^3$, so that $$(9)\qquad 7(x^2 + xy + y^2) - xy(x+y) = 58.$$ Next, let $p=x+y$ and $q = xy$. Then (8) and (9) become $$(10)\qquad 7p-q = 19,\\ (11)\qquad 7(p^2-q) - pq = 58.$$ Substitute the value of $q$ from (10) into (11): $$7p^2 - 7(7p-19) - p(7p-19) = 58,\\ -30p = 58 - 133 = -75,$$ from which $p=\frac52.$ Substitute that into (10) to get $q = -\frac32.$

But $p=x+y$ and $q=xy$. It follows that $x$ and $y$ are the roots of the equation $\lambda^2 - \frac52\lambda -\frac32=0$, namely $3$ and $-\frac12$. So $\{x,y\} = \{3,-\frac12\}$. Substitute those values into (1) and (2) to get $\{a,b\} =\{ 5,16\}$.

Finally, $2014(x+y-xy)-100(a+b) = 2014( p-q) - 100(a+b) = 2014\times4 - 100\times21 = 8056 - 2100 = 5956.$