Solving equations {x + 3 = 5Y, 3x + 2Y + 9 = 0}

Solving equations {x + 3 = 5Y, 3x + 2Y + 9 = 0}

x+3=5y →x=5y-3
Bring in 3x + 2Y + 9 = 0
Then 15y-9 + 2Y + 9 = 0
→y=0
Substituting x = 5y-3
→x=-3

｛x+1=5y+2① ｛3x-9=2y+12②

X = 5Y + 1 (3) is obtained from (1)
3 (5Y + 1) - 9 = 2Y + 12 is obtained by substituting 3 into 2
15y+3-9=2y+12
y=18/13
Substituting y = 18 / 13 into 3 gives x = 103 / 13

Given y = the square of AX + the cube of BX + CX-1, when x = - 2 is y = - 5, find the value velocity of y when x = 2, thank you, thank you
It's not to the fifth power

Wrong, it should be the fifth power of ax
x=-2
y=-32a-8b-2c-1=-5
-(32a+8b+2c)-1=-5
32a+8b+2c=4
So x = 2
y=32a+8b+2c-1=4-1=3