141
№1105.
x + 2x = 18, 3x = 18, x = 6. Ответ: (3; 3).
№1106.
Подставим x и y в уравнение: a⋅2 + 2 ⋅ 1 = 8, 2a + 2 = 8, 2a = 6, a = 3.
Т.е. 3x + 2y = 8.
№1107.
а) 2c(c – 4)
2
– c
2
(2c – 10) = 2c(c
2
– 8c + 16) – 2c
3
+ 10c
2
=
= 2c
3
– 16c
2
+ 32c – 2c
3
+ 10c
2
= –6c
2
+ 32c = 2c(–3c + 16);
при c = 0,2, 2c(–3c + 16) = 2 ⋅ 0,2(–3 ⋅ 0,2 + 16) = 0,4 ⋅ 15,4 = 6,16
б) (a – 4b)(4b + a) = a
2
– 16b
2
;
при a = 1,2; b = –0,6, a
2
– 16b
2
= 1,44 – 16 ⋅ 0,36 = 1,44 – 5,76 = –4,32.
№1108.
а) 1+a–a
2
– a
3
= (1 + a) – a
2
(1 + a) = (1 – a
2
)(1 + a)=(1–a)(1 + a)(1 + a);
б) 8–b
3
+4b–2b
2
=(2–b)(4+2b + b
2
) + 2b(2 – b)=(2–b)(4 + 2b + b
2
+ 2b) =
= (2 – b)(4 + 4b + b
2
) = (2 – b)(2 + b)
2
= (2 – b)(2 + b)(2 + b).
40. График линейного уравнения с двумя переменными
№1109.
3x + 4y = 12;
а) A(4; 1), т.к. 3 ⋅ 4 + 4 ⋅ 1 = 12 + 4 = 16 ≠ 12 , то A ∉ графику;
б) B(1; 3), т.к. 3 ⋅ 1 + 4 ⋅ 3 = 3 + 12 = 15 ≠ 12 , то B ∉ графику;
в) C(–6; –7,5), т.к. 3 ⋅ (–6) + 4 ⋅ (–7,5) = –48 ≠ 12 , то C ∉ графику;
г) D(0; 3), т.к. 3 ⋅ 0 + 4 ⋅ 3 = 12, то D ∈ графику.
№1110.
x – 2y = 4;
а) A(6; 1), т.к. 6 – 2 ⋅ 1 = 4, то A ∈ графику;
б) B(–6; –5), т.к. –6 – 2 ⋅ (–5) = 4 , то B ∈ графику;
в) C(0; –3), т.к. 0 – 2 ⋅ (–2) = 4, то C ∈ графику;
г) D(–1; 3), т.к. –1 – 2 ⋅ 3 = –7 ≠ 4, то D ∉ графику.
№1111.
3x – y = –5, т.к. 3 ⋅ (–1) – 2 = –5, то P ∈ графику;
–x + 10y = 21, т.к. –(–1) + 10 ⋅ 2 = 21, то P ∈ графику;
11x + 21y = 31, т.к. 11 ⋅ (–1) + 21 ⋅ 2 = 31, то P ∈ графику.
№1112.
а) 2x – y = 6
y = 2x – 6
x 0 3
y –6 0
6
3
0
= 2
x
–6