Математика - система

(2x+3)²=5y, (3x+2)²=5y
4x²+12x+9=5y, 9x²+12x+4=5y/*(-1)
4x²+12x+9=5y, -9x²-12x-4=-5y
Сложим их:
4x²-9x²+12x-12x+9-4=5y-5y
-5x²+5=0
5x²=5/:5
x²=1
x=±1
1) x=1
(2*1+3)²=5y
5²=5y/:5
y=5
2) x=-1
(2*(-1)+3)²=5y
(-2+3)²=5y
5y=1/:5
y=0,2
Ответ: (1;5), (-1;0,2)

{ (2x + 3)² = 5y
{ (3x + 2)² = 5y
4x² + 12x + 9 = 5y
9x² + 12x + 4 = 5y
4x² + 12x + 9 = 9x² + 12x + 4
4x² + 12x + 9 - 9x² - 12x - 4 = 0
- 5x² + 5 = 0 | :(-5)
x² - 1 = 0
x² = 1
x = √1 = ±1

(2x + 3)² = 5y при x = 1
(2*1 + 3)² = 5y
5² = 5y
25 = 5y
y = 25/5 = 5

(2x + 3)² = 5y при x = -1
(2*(-1) + 3)² = 5y
1² = 5y
1 = 5y
y = 1/5 = 0,2
Ответ: (1 ; 5) , (-1 ; 0,2)

{ (2x + 3)² = 5y
{ (3x + 2)² = 5y
Правые части равни => равны и левые =>
(2x + 3)^2 = (3x + 2)^2
(2x + 3)^2 - (3x + 2)^2 = 0
[(2x + 3) + (3x + 2)] * [(2x + 3) - (3x + 2] = 0
(5x + 5) * (5 - x) = 0
5*(x + 1) * (5 - x) = 0
(x+1) = 0 -----> x1 = -1
(5-x) = 0 ------> x2 = 5
{ (2x + 3)² = 5y
y = (2x+3)/5
y1 = (2*(-1) + 3)^2/5 = 1/5
y2 = (2*5 + 3)/5 = 13/5
Проверка (при x1=-1; y1 = 1/5:
{ (3x + 2)² = 5y
(3*(-1) + 2)^2 = 5*1/5
1 = 1