Відповіді на вправи: 13.20 (1)

13.20. Розв’яжіть систему рівнянь:

1) x² – 12xy + 36y² = 36, (x – 6y)² = 36, x – 6y = ±6,

x + 6y = 8; x + 6y = 8; x + 6y = 8.

a) x – 6y = –6,

x + 6y = 8; 1 + 6y = 8; 6y = 7; y = $\frac{7}{6}$.

2x = 2, x = 1;

б) x – 6y = 6, +

x + 6y = 8; 7 + 6y = 8; 6y = 1; y = $\frac{1}{6}$.

2x = 14, x = 7;

Відповідь: (1; $\frac{7}{6}$), (7; $\frac{1}{6}$).

2) y² – 2xy = 32, y² – 2xy = 32, y² – 2xy = 32,

x² + 6xy + 9y² = 100; (x + 3y)² = 100; x + 3y = ±10.

a) x = –3y – 10, x = –3y – 10,

y² – 2y(–3y – 10) – 32 = 0; y² + 6y² + 20y – 32 = 0;

x = –3y – 10,

7y² + 20y – 32 = 0.

7y² + 20y – 32 = 0; D = 400 + 896 = 1296 = 36²;

y₁ = $\frac{-20-36}{14}$ = $\frac{-56}{14}$ = –4; y₂ = $\frac{-20+36}{14}$ = $\frac{16}{14}$ = $\frac{8}{7}$;

x₁ = –3 • (–4) – 10 = 2; x₂ = –3 • $\frac{8}{7}$ – 10 = –$\frac{24}{7}$ – 10 = –3$\frac{3}{7}$ = –$\frac{94}{7}$.

б) x = 10 – 3y, x = 10 – 3y,

y² – 2y(10 – 3y) = 32; y² – 20y + 6y² – 32 = 0;

x = 10 – 3y,

7y² – 20y – 32 = 0.

7y² – 20y – 32 = 0; D = 400 + 896 = 1296; y₁ = $\frac{20-36}{14}$ = –$\frac{16}{14}$ = –$\frac{8}{7}$;

y₂ = $\frac{20+36}{14}$ = $\frac{56}{14}$ = 4. x₁ = 10 + 3 • $\frac{8}{7}$ = $\frac{94}{7}$; x₂ = 10 – 3 • 4 = –2.

Відповідь: (2; –4), (–$\frac{94}{7}$; $\frac{8}{7}$), ($\frac{94}{7}$; –$\frac{8}{7}$), (–2; 4).

3) x² + y² = 25, x² + y² = 25, +

xy = 12; • 2 2xy = 24; x + y = 7 або x + y = –7.

(x + y)² = 49;

a) x + y = 7, x = 7 – y, x = 7 – y, x = 7 – y,

xy = 12; y(7 – y) = 12; y² – 7y + 12 = 0; y = 3 або y = 4;

x = 4, або x = 3,

y = 3 y = 4.

б) x + y = –7, x = –y – 7, x = –y – 7,

xy = 12; y(–y – 7) –12 = 0; y² + 7y + 12 = 0;

x = –y – 7, y = –3, або y = –4,

y = –3 або y = –4; x = –4 x = –3.

Відповідь: (4; 3), (3; 4), (–4; –3), (–3; –4).

4) 9x² + y² = 10, 9x² + y² = 10, +

xy = –1; • 6 6xy = –6;

(3x + y)² + 4;

3x + y = 2 або 3x + y = –2.

a) 3x + y = 2, y = 2 – 3x, y = 2 – 3x,

xy = –1; x(2 – 3x) + 1 = 0; 2x – 3x² + 1 = 0;

y = 2 – 3x, y = 2 – 3x, x = 1, або

3x² – 2x – 1 = 0; x = 1 або x = –$\frac{1}{3}$; y = 2 – 3 • 1

x = –$\frac{1}{3}$, x = 1, або x = –$\frac{1}{3}$,

y = 2 + 3 • $\frac{1}{3}$; y = –1 y = 3.

б) 3x + y = –2, y = –3x – 2, y = –3x – 2,

xy = –1; x(–3x – 2) + 1 = 0; 3x² + 2x – 1 = 0;

y = 2 – 3x, x = –1, або x = $\frac{1}{3}$,

x = –1 або x = $\frac{1}{3}$; y = 1 y = –3.

Відповідь: (1; –1), (–$\frac{1}{3}$; 3), (–1; 1), ($\frac{1}{3}$; –3).