Решите систему уравнений:

а) (х – 2) ⋅ (у + 3) = 160, у – х = 1

б) (х – 1) ⋅ (у + 10) = 9, х – у = 11.

Решение:

а) (х – 2) ⋅ (у + 3) = 160

у – х = 1

1) y – x = 1

y = x + 1

2) (x – 2) (x + 1 + 3) = 160

(x – 2) (x + 4) = 160

x² – 2x + 4x – 8 – 160 = 0

x² + 2x – 168 = 0

D = 2² – 4 ⋅ 1 ⋅ (-168) = 4 + 672 = 676 > 0

Есть 2 корня

x₁ = -2 + √ 676 / 2 = -2 + 26 / 2 = 24 / 2 = 12

x₂ = -2 – √ 676 / 2 = -2 – 26 / 2 = -28 / 2 = – 14

3) При x = 12

y = 12 + 1 = 13

При x = -14

y = – 14 + 1 = -13

(12;13) , (-14;-13)

б) (х – 1) ⋅ (у + 10) = 9

х – у = 11.

1) x – y = 11

y = x – 11

2) (x – 1) (x – 11 + 10) = 9

(x – 1) ( x – 1) = 9

(x – 1)² – 9 = 0

x² – 2x + 1 – 9 = 0

x² – 2x – 8 = 0

D = (-2)² – 4 ⋅ 1 ⋅ (-8) = 4 + 32 = 36 > 0

Имеется 2 корня

x₁ = 2 + √ 36 / 2 = 2 + 6 / 2 = 8 / 2 = 4

x₂ = 2 – √ 36 / 2 = 2 – 6 / 2 = -4 / 2 = – 2

3) При x = 4

y = 4 – 11 = -7

При x = -2

y = -2 – 11 = -13

(4; -7), (-2; -13)

Источник : Учебник по Алгебре 9 класса авторов Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова, 2014г.