# System of Higher Degree Equations on Two Variables

## Algebra Math Problem: How to solve the system of higher degree equations on two variables

Find the real roots of the system of higher degree equations, **x ^{4} + y^{4} = 82** and

**x – y = 2**

## Solution

Let

x^{4} + y^{4} = 82………..eq(1)

x – y = 2……………..eq(2)

assume x = u + v and y = u – v, then

From equation 2

x – y = u + v – (u – v)

⇒ 2 = u + v – u + v

⇒ 2 = 2v

so, v = 1

From equation 1

x^{4} + y^{4} = 82

⇒ 82 = (u + v)^{4} + (u – v)^{4}

substitute v = 1

82 = (u + 1)^{4} + (u – 1)^{4}

simplify the equation

82 = u^{4} + 4u^{3} + 6u^{2} + 4u + 1 + (u^{4} – 4u^{3} + 6u^{2} – 4u + 1)

⇒ 82 = 2u^{4} + 12u^{2} + 2

⇒ 2u^{4} + 12u^{2} – 80 = 0

thus, u^{4} + 6u^{2} – 40 = 0

## This is a quadratic equation with u^{2}, We can solve this equation with the use of factorization

u^{4} + 6u^{2} – 40 = u^{4} + 10u^{2} – 4u^{2} – 40

⇒ 0 = u^{2}(u^{2} + 10) – 4(u^{2} + 10)

⇒ 0 = (u^{2} + 10)(u^{2} – 4)

From the equation we get

u^{2} + 10 = 0 or u^{2} – 4 = 0

⇒ u^{2} = 10 or u^{2} = 4

that is, u = ±√-10 or u = ±√4

When u = ±√-10 x has no real roots

so, u = ±2

When u = 2, then

x = u + v = 2 + 1 = 3

y = u – v = 2 – 1 = 1

When u = – 2, then

x = u + v = – 2 + 1 = -1

y = u – v = – 2 – 1 = -3

then solution to the system of higher degree equations are **(3, 1), ( – 1, – 3)**