Question 1151472
<pre>
There may be some trick that gives an easier way than to solve for all the
letters, but here it is that way: 

{{{system(ax + by = 5,
ax^2 + by^2 = 10,
ax^3 + by^3 = 50,
ax^4 + by^4 = 130)}}}
Solve the first two equations for a, by Cramer's rule:
{{{system(ax^"" + by^"" = 5,
ax^2 + by^2 = 10)}}}
{{{a=abs(matrix(2,2,5,y,10,y^2))/abs(matrix(2,2,x^"",y^"",x^2,y^2))=
(5y^2-10y)/(xy^2-x^2y)=(5y(y-2))/(xy(y-x))=5y(y-2)/(xy(y-x))=(5(y-2))/(x(y-x))
)}}}
Solve the 2nd and 3rd equations for a, by Cramer's rule:
{{{system(ax^2 + by^2 = 10,
ax^3 + by^3 = 50)}}}
{{{a=abs(matrix(2,2,10,y^2,50,y^3))/abs(matrix(2,2,x^2,y^2,x^3,y^3))=
(10y^3-40y^2)/(x^2y^3-x^3y^2)=10y^2(y-5)/(x^2y^2(y-x))=(10^""(y-5))/(x^2(y-x))
)}}}
Set the first and second values for a equal:
{{{(5(y^""-2))/(x^""(y^""-x))=(10(y^""-5))/(x^2(y^""-x))}}}
Multiply both sides by
{{{(x(y-x))/5}}}
{{{y-2=2(y-5)/x}}}
{{{x(y-2)=2(y-5)}}}
equation 1:    {{{x=(2(y-5))/(y-2)}}}
-----------------------------
Solve the 3rd and 4th equations for a, by Cramer's rule:
{{{system(ax^3 + by^3 = 50,
ax^4 + by^4 = 130)}}}
{{{a=abs(matrix(2,2,50,y^3,130,y^4))/abs(matrix(2,2,x^3,y^4,x^4,y^4))=
(50y^4-130y^3)/(x^3y^4-x^4y^3)=(10y^3(5y-13))/(x^3y^3(y-x))=(10(5y-13))/(x^3(y-x))
)}}}
Set the second and third values for a equal:
{{{(5(y^""-2))/(x^""(y^""-x))=(10(y^""-5))/(x^2(y^""-x))}}}
Multiply both sides by
{{{(x^2(y-x))/10^""}}}
{{{y-3=(5y-13)/x}}}
{{{x(y-3)=5y-13}}}
equation 2:    {{{x=(5y-13)/(y-5)}}}
Set the values of x from equations 1 and 2 equal
{{{(2(y-5))/(y-2)=(5y-13)/(y-5)}}}

{{{2(y-5)^2=(y-2)(5y-13)}}}

{{{2(y^2-10y+25)=5y^2-23y+26}}}
{{{2y^2-20y+50=5y^2-23y+26}}}
{{{0=3y^2-3y-24}}}
{{{0=y^2-y-8}}}
{{{y = (-(-1) +- sqrt((-1)^2-4(1)(-8) ))/(2(1)) }}}
{{{y = (1 +- sqrt(1+32 ))/2 }}}
{{{y = (1 +- sqrt(33))/2 }}}

x and y cannot be equal because the first two given equations
would be inconsistent. For if they were equal, we'd have
{{{ax+bx=5}}}
Nultiplying thru by x gives
{{{ax^2+bx^2=5x=50}}} or that would give x=10, not what we got for y.
Therefore by symmetry of x and y, one of them has the + sign and the
other has the minus sign. So we choose them so that

{{{y = 1/2 + sqrt(33)/2 }}}
and
{{{x = 1/2 - sqrt(33)/2 }}}

Since
{{{a=(10^""(y-5))/(x^2(y-x))
)}}}
{{{y-5=1/2 + sqrt(33)/2-5=-9/2+sqrt(33)/2}}}, so
{{{10^""(y-5)=10(-9/2+sqrt(33)/2)=-45+5sqrt(33)}}}

{{{y-x=(1/2 + sqrt(33)/2)-(1/2 - sqrt(33)/2)=sqrt(33)}}}

{{{x^2=(1/2 - sqrt(33)/2)^2 = 1/4 - sqrt(33)/2+33/4 = 34/4-sqrt(33)/2=17/2-sqrt(33)/2}}}, so
{{{x^2(y-x)=(17/2-sqrt(33)/2)(sqrt(33))=17sqrt(33)/2-33/2}}}
{{{a=((-45+5sqrt(33))/1)/(17sqrt(33)/2-33/2)=2(-45+5sqrt(33))/(17sqrt(33)-33)
}}}
Rationalizing the denominator we get
{{{a=(5(11 - 5sqrt(33)))/176}}}
By symmetry, or by substituting,
{{{b=(5(11 + 5sqrt(33)))/176}}} 

So

{{{13(x + y - xy) - 120(a + b)}}}

{{{13((1/2 - sqrt(33)/2) + (1/2 + sqrt(33)/2) - (1/2 - sqrt(33)/2)(1/2 + sqrt(33)/2)) - 120(((5(11 - 5sqrt(33)))/176) + ((5(11 + 5sqrt(33)))/176))}}} 

{{{13(1/2 - sqrt(33)/2 + 1/2 + sqrt(33)/2 - (1/2 - sqrt(33)/2)(1/2 + sqrt(33)/2)) - 120((5(11 - 5sqrt(33)))/176 + (5(11 + 5sqrt(33)))/176)}}}

{{{13(1/2 - sqrt(33)/2 + 1/2 + sqrt(33)/2 - (1/4 - 33/4)) - 120((5(11 - 5sqrt(33)))/176 + (5(11 + 5sqrt(33)))/176)}}}

{{{13(1 - (-32/4)) - 120((55 - 25sqrt(33))/176 + (55 + 25sqrt(33))/176)}}}

{{{13(1 - (-8)) - 120(((55 - 25sqrt(33))+ (55 + 25sqrt(33)))/176)}}}

{{{13(1+8) - 120((55 - 25sqrt(33)+ 55 + 25sqrt(33))/176)}}}

{{{13(9) - 120(110)/176)}}}

{{{117 - 13200/176)}}}

{{{117 - 75)}}}

{{{42}}}

Edwin</pre>