Հարցերի քննարկում:
Թեմա՝ Երկու անհայտով առաջի աստիճանի երկու հավասարումների համակարգեր, տեղադրման եղանակ:
Ա․ x + 2y-3 = 0, x = 0 + 3 + 2y = 3 + 2y
3 + 2y + y + 1 = 0
2y + y = 3y
3y = 3 + 1 = 4
x = 4
4 + 2y + 3 = 9
y = 9
(9;4)
Բ․ x-3y + 3 = 0, x = 0 + 3y-3 = 3y-3
3y-3 + y-1 = 0
3y + y = 4y
4y = 3 + 1 = 4
x = 4
4 + 3y-3 = 4
y = 4
(4; 4)
Գ․ 4x + y-2 = 0, y = 0 + 2 + 4x = 2 + 4x
3x + 2 + 4x + 3 = 0
3x + 4x = 7x
7x = 2 + 3 = 5
y = 5
5 + 2 + 4x = 11
x = 11
(11; 5)
Դ․ xy-7 = 0, x = 0 + 7 + y = 7 + y
3 . (7 + y) – y + 1 = 0
21 + 3y-y + 1 = 0
2y = 22
22: 2 = 11
y = 11
11 + 7 = 18
x = 18
(18; 11)
Ա․ x + 3y-1 = 0, x = 0 + 1 + 3y = 1 + 3y
-1 + 3y + 4y + 8 = 0
-3y-4y = -7
-7 = 1 + 8 = 9
x = 9
9 + 3y + 1 = 13
y = 13
(9; 13)
Բ․ x-2y + 3 = 0, x = 0 + 3-2y = 3-2y
-3-2y + 3y-2 = 0
-2y-3y = -5
-5 = 3 + 2 = 5
x = 5
5 + 2y-3 = 4
y = 4
(5; 4)
Գ․ x-y + 2 = 0, x = 0 + 2-y = 2-y
3. (2-y) + y – 4 = 0
3y-6 + y-4 = 0
3y + y = 4y
-6-4 = -10
10 = 4y
y = 2,5
x = 0,5
(0,5;2,5)
Դ․ 2x + y-3 = 0, y = 0 + 3 + 2x = 3 + 2x
-x. (3 + 2x) + 4 = 0
-2x – (- 3) + 4 = 0
-2x – (- 3) + 4 = 1 = 5
5 = 5
y = 1
1 + 2x + 3 = 6
x = 6
(6; 1)
Ա․ x + 2y-3 = 0, x = 0 + 3 + 2y = 3 + 2y
2. (3 + 2y) – 3y + 8 = 0
4y + 6-3y + 8 = 0
4y-3y = 1y
6 + 8 = 14
14 = 1y
x = 1
1 + 2y + 3 = 6
y = 6
(1; 6)
Բ․ 2x + y-8 = 0, y = 0 + 8 + 2x = 8 + 2x
3x + 4 . (8 + 2x) – 7 = 0
3x + 8x + 32-7 = 0
3x + 8x = 11y
32-7 = 25
25 = 11y
y = 2,27
2.27 + 2x + 8 = 12,27
x = 12,27
(12,27; 2,27)
Գ․ -6x+2y+6=0, x=-6x=6-2y
x=6-2y:(-6)=6,3
5x . (6,3x) – y – 17=0
5x . 6,3=31,5
31,5=17-y=16
x=16
16+6-2=20
y=20
(16;20)
Դ․ 5x+3y-7=0, 3y=7+5x
2x – (7+5x) – 5=0
2x-5x=-3x
-3x=7-5=2
y=2
2+7+5x=14
x=14
Արփինե Խաչատրյանի բլոգ 