(3y-1)²=(y-3)²+2y² Don't use an open root

(3y-1)²=(y-3)²+2y² Don't use an open root

9y²-6y+1=y²-6y+9+2y²
6y²=8
y²=4/3
y=-2√3/3,y=2√3/3

2 | X-Y | + √ 2Y + Z + Z & sup2-z + 1 / 4 = 0 find the value of X + y + Z

Y = root two of minus eight, x = root two of minus eight, z = 0
X + y + Z = negative quarter root two

The system of equations {x-y-1 = 0,4 (X-Y) - y = 5 is brought into the normal solution {2x-3y-2 = 0,2x-3y + 5 / 7 + 2Y = 9} by multiplying X-Y = 1,4 by 1-y-5 of 1, y = - 1 as a whole

2x-3y = 2, bring in (2)
2+5/7+2Y=9
Y=22/7
X=40/7

Solving the equation y & sup2; - 7 / 2Y + 3 = 0 by substitution method

y²-7/2y+3=0
（y-1.5)(y-2)=0
y1=1.5 y2=2

Solve the equations: x + 13 = 2Y2 (x + 1) − y = 11

Solution (1): from the original system of equations to get x = 6y − 12x − y = 9, substitute ① into ② to get 2 (6y-1) - y = 9, i.e. y = 1; substitute ① to get x = 5; the solution of the original system of equations is x = 5Y = 1. Solution (2): from x + 13 = 2Y to get x + 1 = 6y, substitute ① into 2 (x + 1) - y = 11 to get 12y-y = 11, i.e. y = 1; substitute y = 1 into ①

The solution of 1 / 2x-3 / 2Y = - 1 2x + y = 3
This is a problem of quadratic equation of two variables!

1 / 2x-3 / 2Y = - 1, then x-3y = - 2, then x = 3y-2, substituting 2x + y = 3, we get
6y-4 + y = - 2 + 9, the solution is y = 1
Substituting x = 3y-2, the solution is x = 1

Factorization by Collocation: x ^ 2 + 2x-3 2x ^ 2-7x + 6 3x ^ 2Y ^ 2-10xy + 7 - y ^ 2 + 20y-96

x^2+2x-3=(x+3)(x-1)
2x^2-7x+6=(x-2)(2x-3)
3x^2y^2-10xy+7=(xy-1)(3xy-7)
-y^2+20y-96=-(y-8)(y-12)

Use proper method to solve the following equation: the square of 2 / 3 y + 1 / 3 Y-2 = 0
Solve the equation: the fourth power of X - the square of X - 6 = 0

If you multiply both sides by 3, you get,
2Y^2+Y-6=0,
Cross multiplication
（2y-3)(y+2)＝0
We get Y1 = 1.5, y2 = - 2
The fourth power of X - the square of X - 6 = 0
（X^2-3）（X^2+2）=0
(x ^ 2 + 2) constant greater than 0
(x ^ 2-3) = 0, x = ± root 3

The formula method is used to solve the following equation
I'd like to ask how 1 / 4 and 1 of them come from? Who can help? It needs a clear process

x²－x－3/4＝0
x²－x＝3/4
x²－x﹢(1/2)²=3/4＋1/2
(x－1/2)²＝5/4
X-1 / 2 = root of plus or minus 2 / 5
Ψ X1 = root of two 5 + 1 x2 = root of negative two 5 + 1

-How to calculate the equation of Y square + 2Y + 3 = 0

-The square of Y + 2Y + 3 = 0
Can be transformed into
-(y²-2y-3)=0
That is - (Y-3) (y + 1) = 0
What is the root of
y1=3,y2=-1