Solving the equation y - [3y-1] is equal to 5 [y + 2]

Solving the equation y - [3y-1] is equal to 5 [y + 2]

y-[3y-1]=5【y+2]
y-3y+1=5y+10
3y=-9
y=-3

Y + 17 divided by 5 minus 3y-7 divided by 4 equals to - 2 for y

Take 20 on both sides
4(y+17)-5(3y-7)=-40
4y+68-15y+35=-40
15y-4y=68+35+40
11y=143
y=143÷11
y=13

Solving equation (1) the square of x minus 2XY minus y equals 1 (2) the square of 2x minus 5xy minus 3Y equals 0

(2) The formula can be decomposed into (2x + y) (x-3y) = 0, and there are two possibilities for multiplication to be zero: 1.2x + y = 0, 2.x-3y = 0
Let's first look at the first one, which may be derived from 2x + y = 0, where y = - 2x is brought into (1), where x = 1, y = - 2, or x = - 1, y = 2
The second possibility is that x = 3Y can be obtained from x-3y = 0, y square = 1 / 2 can be obtained from (1), and then x can be obtained