Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 387

1. 6(у – х) – 50 = у
6у – 6х – 50 – у = 0
5у = 6х + 50
у = 6х + 50/5
у = 1,2х + 10

2. 1,2х + 10 – х(1,2х + 10) = 24
1,2х + 10 – 1,2х² – 10х – 24 = 0
–1,2х² – 8,8х – 14 = 0 | · (–2,5)
3х² + 22х + 35 = 0
D = 22² – 4 · 3 · 35 = 484 – 420 = 64

х₁ = $$\frac{-22 + \sqrt{64}}{2 \cdot 3}$$ = –22 + 8/6 = –14/6 = –7/3 = –21/3

х₂ = $$\frac{-22 - \sqrt{64}}{2 \cdot 3}$$ = –22 – 8/6 = –30/6 = –5

3. х = –21/3

y = 1,2 · (–21/3) + 10 = –12/10 · 7/3 + 10 = –2,8 + 10 = 7,2

x = –5

у = 1,2 · (–5) + 10 = –6 + 10 = 4

Ответ: (–21/3; 7,2); (–5; 4)

1. p + 5t = 2(p + t)
p + 5t = 2p + 2t
5t – 2t = 2p – p
3t = p

2. t · 3t – t = 10
3t² – t – 10 = 0
D = (–1)² – 4 · 3 · (–10) = 1 + 120 = 121

t₁ = $$\frac{1 + \sqrt{121}}{2 \cdot 3}$$ = 1 + 11/2 · 3 = 12/6 = 2

t₁ = $$\frac{1 - \sqrt{121}}{2 \cdot 3}$$ = 1 – 11/2 · 3 = –10/6 = –5/3 = –12/3

t = 2
p = 2 · 3 = 6
t = –12/3
p = 3 · (–12/3) = 3 · (–15/3) = –5

Ответ: (6; 2), (–5; –12/3)