ALjabar

ax + by = 4

ax^2 + by^2 = 16

ax^3 + by^3 = 32

ax^4 + by^4 = 80

(ax + by)(x + y) = ax² + by² + xy(a + b)

4(x + y) = 16 + xy(a + b)

4(x + y) - xy(a + b) = 16 ... (m)

(ax² + by²)(x + y) = ax³ + by³ + xy(ax + by)

16(x + y) - 4xy = 32

4(x + y) - xy = 8 ... (a)

(ax³ + by³)(x + y) = ax⁴ + by⁴ + xy(ax² + by²)

32(x + y) = 80 + 16xy

4(x + y) - 2xy = 10 ... (t)

pers (a) dan (t)

4(x + y) - xy = 8

4(x + y) - 2xy = 10__(-)

xy = -2 ... (h)

4(x + y) - xy = 8

x + y = (8 + xy)/4

x + y = (8 - 2)/4

x + y = 3/2

2x + 2y = 3 ... (g)

pers (h) dan (g)

xy = -2

x (2y) = -4

x(3 - 2x) = -4

2x² - 3x - 4 = 0

rumus abc

x1.2 = (-b ± √(b² - 4ac))/2a

x1.2 = (3 ± √41)/4

• solusi_1

x = (3 + √41)/4

y = 3/2 - x

y = (3 - √41)/4

4(x + y) - xy(a + b) = 16

4 × 6/4 + 2(a + b) = 16

a + b = (16 - 6)/2

a + b = 5 ... (i)

ax + by = 4

a(3 + √41)/4 + b(3 - √41)/4 = 4

3a + a√41 + 3b - b√41 = 16

3(a + b) + (a - b)√41 = 16

(a - b)√41 = 16 - 15

a - b = 1/√41 ... (c)

pers (i) dan (c)

a + b = 5

a - b = 1/√41__(+)

a = (5 + 1/√41)/2

b = (5 - 1/√41)/2

• solusi_2

dg cara yg sama

x = (3 - √41)/4

y = (3 + √41)/4

a = (5 - 1/√41)/2

b = (5 + 1/√41)/2