Solving the equations (x + y) / 6 - (X-Y) / 4 = 0 (2x-y) / 3 + (x-2y) / 2 = 9, x = y= Why, Title Format (x+y)/6 -(x-y)/4=0 (2x-y)/3+(x-2y)/2=9 x= y=

Solving the equations (x + y) / 6 - (X-Y) / 4 = 0 (2x-y) / 3 + (x-2y) / 2 = 9, x = y= Why, Title Format (x+y)/6 -(x-y)/4=0 (2x-y)/3+(x-2y)/2=9 x= y=

(x + y) / 6 - (X-Y) / 4 = 0 is 5y-x = 0, that is 35y-7x = 0
(2x-y) / 3 + (x-2y) / 2 = 9, namely 7x-8y = 54
The sum of the two formulas is 27y = 54
y=2 x=10

(x + y) z = 1 (1.2x + y) 9 / 10 = 1 (x + 3 / 4Y) (Z + 2.5) = 1

(1.2X+Y)9/10=1,10.8X+9Y=10,9Y=10-10.8X
Put 9y into 1,3
(10-1.8X)Z=9
(1.2X+10)(Z+2.5)=12
Eliminate x, 50z? 2 + 71z-45 = 0
Solve the quadratic equation of one variable

To solve the equations: x square - 2XY + y square = 1 and 2x square - 5xy - 3Y square = 0

This is not very difficult:
First of all, according to the first equation, x ^ 2-2xy + y ^ 2 = (X-Y) ^ 2 = 1, we get the following results:
X-Y = 1 or X-Y = - 1
Then, the calculation is divided into two cases
(1) If x = y + 1 is substituted into the second equation, 2x ^ 2-5xy-3y ^ 2 = 0, we can get a quadratic equation of one variable about y
(2) If x = Y-1 is substituted into the second equation, y, X can be solved
In this way, we can sort out four groups of solutions
It's better to arrange and calculate by yourself. This is not a big problem~

In this paper, we first simplify and then evaluate the square of X / x-2xy + y square of x times the square of X-Y / x + y + (2x + 2 / X-Y) - 2, where x, y satisfy the equation system x + 2Y = 3, X-Y = - 5

This topic is a bit confusing If it is understood as [x / (x? - 2XY + y]] * [(x? - y?) / (x + y)] + (2x + 2) / (X-Y) - 2, then it can be simplified as follows: = [x / (X-Y) 2] * [(x + y) * (X-Y) / (x + y)] + (2x + 2) / (X-Y) - 2 = x / (X-Y) + (2x + 2) / (X-Y) - 2 = [x + (2x + 2) - 2 (X-Y)] / (X -...)

Solving equations y = x + 1 ① y = x squared + 2x-1 ②

y=x+1①
y=x²+2x-1②
Substituting y = x + 1 into (2)
x+1=x²+2x-1
x²+x-2=0
Cross multiplication
1 -1
12
(x-1)(x+2)=0
X = 1 or x = - 2
therefore
Y = 1 + 1 = 2 or y = - 2 + 1 = 1
therefore
X = 1, y = 2 or
x=-2 y=1

The equation system, x square - XY + 2x = 0, xsquare - 2XY + y + 8 = 0,

If x = 0, then x = - 8, x = 0, y = - 8, if x ≠ 0, then X-Y + 2 = 0, then x = 0 or X-Y + 2 = 0. If x = 0, then x = - 8, x = 0, y = - 8. If x ≠ 0, then X-Y + 2 = 0, that is, x = Y-2. If x? - 2XY + y + 8 = 0, then y? - 12 = 0, then y = - 3 or y = 4. If y = 4, then x = 2, if y = - 3, then x =

Given the equations 4x-3y = 6Z 2x + 4Y = 14z, calculate the value of x square + y square / 2XY

From the equations 4x-3y = 6Z 2x + 4Y = 14z, x = 3Z, y = 2Z; substituting x square + y square / 2XY, we get 13 / 12

To solve the equations: (x + y) ^ 2-2xy - (x ^ 2) (y ^ 2), x + y-xy = 4 Solving equations (x+y)^2-2xy-(x^2)(y^2)=10 x+y-xy=4

Let a = x + y, B = XY equations be changed into: a? - 2b-b? = 10a-b = 4, replace a = B + 4 into Formula 1: B? + 8b + 16-2b-b? = 10, then B = - 1, thus a = 3, that is, x + y = 3, xy = - 1. Therefore, x, y are the two root solutions of the equation Z? - 3z-1 = 0, and z = (3 ±√ 13) / 2. Therefore, the original equation has two sets of solutions: x = (3 + √ 13)

Solving equations: XY / x + y = - 6 (1) 2XY / x + 2Y = 12 (2)

Hello:
XY
——=-6①
X+y
2XY
——=12②
x+2y
xy=-6（x+y）③
2xy=12（x+2y）④
③×2
2xy=-12（x+y）
Then - 12 (x + y) = 12 (x + 2Y)
-12x-12y=12x+24y
-24x=36y
x=-1.5y⑤
-1.5y^2=-15y
y=10
x=10×(-1.5)=-15

Solving equations: 1, x ^ 2-2xy-y ^ 2 = 1 2, under the radical sign (XY + 3) = X

√(xy+3)=x
xy+3=x^2
So y = (x ^ 2-3) / X
Substitute x ^ 2-2xy-y ^ 2 = 1
x^2-2(x^2-3)-(x^2-3)^2/x^2=1
-x^2+6-(x^2-3)^2/x^2=1
-x^4+6x^2-x^4+6x^2-9=x^2
2x^4-11x^2+9=0
(2x^2-9)(x^2-1)=0
x^2=1,x^2=9/2
x=±1,x=±3√2/2
y=(x^2-3)/x
So x = 1, y = - 2
x=-1,y=2
x=3√2/2,y=√2/2
x=-3√2/2,y=-√2/2
The test shows that x is less than 0
So x = 1, y = - 2
Or x = 3 √ 2 / 2, y = √ 2 / 2