# Challenge Problem 12: Tricky Algebra Problem From Nepal Olympiad

Suppose:
ax + by = 5
ax^2 + by^2 = 10
ax^3 + by^3 = 50
ax^4 + by^4 = 130
Calculate the value of:
13(x + y – xy) – 120(a + b)

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This problem is adapted from the Mathematics Olympiad of Nepal (MOON). Watch the video for a solution.

We need to evaluate the expression:

13(x + y – xy) – 120(a + b)

Rather than solving for each variable, it will be sufficient to solve for these values:

x + y
xy
a + b

The key is the equations are related to each other if we multiply by x + y. To see how, start with the degree 2 equation:

ax^2 + by^2 = 10

Multiply both sides by x + y to get:

(ax^2 + by^2)(x + y) = 10(x + y)
(ax^3 + by^3) + xy(ax + by) = 10(x + y)

Now substitute ax^3 + by^3 = 50 and ax + by = 5 to get:

50 + 5xy = 10(x + y)

We will do a similar calculation for the equation of degree 3.

ax^3 + by^3 = 50

Multiply both sides by x + y to get:

(ax^3 + by^3)(x + y) = 50(x + y)
(ax^4 + by^4) + xy(ax^2 + by^2) = 50(x + y)

Now substitute ax^4 + by^4 = 130 and ax^2 + by^2 = 10 to get:

130 + 10xy = 50(x + y)

We now have the system of equations:

130 + 10xy = 50(x + y)
50 + 5xy = 10(x + y)

Now multiply the second equation by 2:

130 + 10xy = 50(x + y)
100 + 10xy = 20(x + y)

Then subtract from the first equation to eliminate the term xy, giving:

30 = 30(x + y)
x + y = 1

Substitute x + y = 1 into either of the original equations and we can solve for xy:

130 + 10xy = 50(x + y)
130 + 10xy = 50
xy = -8

Now we will solve for a + b from the first degree equation:

ax + by = 5

Multiply both sides by x + y to get:

(ax + by)(x + y) = 5(x + y)
(ax^2 + by^2) + xy(a + b) = 5(x + y)

Now substitute ax^2 + by^2 = 10, xy = -8, and x + y = 1 to get:

10 – 8(a + b) = 5
a + b = 5/8

Finally we can substitute the solved values to get the answer:

13(x + y – xy) – 120(a + b)
= 13(1 – (-8)) – 120(5/8)
= 13(9) – 75
= 42

References

Problem from the Mathematics Olympiad of Nepal (MOON) emailed to me

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