How to calculate the addition, subtraction and elimination methods of quadratic equations {2x + 3Y = - 1 and {4x-y = 5

How to calculate the addition, subtraction and elimination methods of quadratic equations {2x + 3Y = - 1 and {4x-y = 5

{2x+3y=-1 (1)
{4x-y=5 (2)
(1) 4X + 6y = - 2 (3)
(3) 2
7y=-7
The solution is: y = - 1, substituting (1) to get:
2x-3=-1
2x=2
The solution is: x = 1
So the solution of the original equations is:
{ x=1
{ y=-1

The integer solution of the system of inequalities-3 less than or equal to 2x-1 less than 1 is

Solving inequality-3 ≤ (2x-1) / 3 ＜ 1
It is found that: - 4 ≤ x ＜ 2
The integer solutions are - 4, - 3, - 2, - 1, 0, 1
Answer: the integer solutions of inequality system - 3 less than or equal to 2x-1 of 3 less than 1 are - 4, - 3, - 2, - 1, 0, 1

Calculation: 1 / (X-Y) - 1 / (x + y) - 2Y / (x ^ 2 + y ^ 2) - 4Y ^ 3 / (x ^ 4 + y ^ 4) - 8x ^ 7 / (x ^ 8 + y ^ 8)

1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=【（x+y)-(x-y)】/(x-y)(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=2y/(x^2-y^2)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=【2y(x^2+y^2）...

Calculation (2x-4y) (2Y + x)

(2X-4Y)(2Y+X)
=2 (x-2y) (x + 2Y)
=2x²-8y²
Do not understand can ask oh, hope to adopt, thank you

Fast calculation (x-2y) (x + 4Y)

Solution (x-2y) (x + 4Y)
=x^2+4xy-2xy-8y^2
=x^2+2xy-8y^2.

Mathematics problem of grade one in junior high school (x + 2Y) ^ 2 (x ^ 2 + 4Y ^ 2) ^ 2 (x-2y) ^ 2 - [x ^ 4 - (2Y ^ 4) ^ 2] ^ 2
(x + 2Y) ^ 2 (x ^ 2 + 4Y ^ 2) ^ 2 (x-2y) ^ 2 - [x ^ 4 - (2Y ^ 4) ^ 2] ^ 2 steps

(2Y ^ 4) ^ 2 should be (4Y ^ 2) ^ 2
Original formula = [(x + 2Y) (x-2y)] ^ 2 (x ^ 4 + 4Y ^ 2)] ^ 2 - (x ^ 4-16y ^ 4) ^ 2
=(x^2-4y^2)^2(x^4+4y^2)]^2-(x^4-16y^4)^2
=(x^4-16y^4)^2-(x^4-16y^4)^2
=0

If x-2y = 7, x + 2Y = 5, then the value of x ^ 2-4y ^ 2 is

(1)x^2-4y^2
=x^2-(2y)^2
=(x+2y)(x-2y)
∵x-2y=7,x+2y=5
=5*7=35
(2)x^2+1/x^2=x^2+1/x^2+2*x*1/x-2
=(x+1/x)^2-2
∵x+1/x=3
=3^2-2=7

0.4y + 09 / 0.5-y-5 / 2 = 0.3 + 0.2y/0.3 (solving equation)

0.6y-2y/3=2.8-9/5
9y-10y=42-13
y=-29
How do you know?

The solution equation (system) is (1) y + 24-2y-16 = 1 (2) 3x-4y = 105x + 6y = 42

(1) If the denominator is removed, 3 (y + 2) - 2 (2y-1) = 12, if the bracket is removed, 3Y + 6-4y + 2 = 12, if the term is shifted, 3y-4y = 12-6-2, if the similar term is combined, y = 4, the coefficient of X is reduced to 1, y = - 4; (2) 3x-4y = 10, ① 5x + 6y = 42, ②, ① × 3, ② × 2, 9x-12y = 30, ③ 10x + 12Y = 84, ④, ③ +

-2Y + 3-y = 2-4y-1

-3y+4y=2-1-3
y=-2