How can a quadratic equation of two variables be transformed into a quadratic equation of one variable? Let's put it in common

How can a quadratic equation of two variables be transformed into a quadratic equation of one variable? Let's put it in common

There should be two formulas for the quadratic equation of two variables (for example, x + y = 5 and 5x + 2Y = 16)
Let's change the other side of the equation from x = 5 to y + 5
Then, the transformed formula is transformed into another unchanged equation [for example, if x = 5-y is replaced by 5x + 2Y = 16, it becomes
5＊（5－Y）＋2Y＝16〕
And then the equation becomes a linear equation
(for example, the above formula becomes 25-5y + 2Y = 16
3Y＝9
Y＝3）
In the end, just find out X
One of the unknowns is expressed by an equation containing the other unknowns, and then brought into another equation
Bring in elimination method: ① x + y = n, ② 2x + 2Y = m, we know that x = M-Y, bring in ② to get n-y + 2Y = m (m.n is constant)
The addition and subtraction elimination method can be used: one formula has ① x + (-) ny, and the other formula ② x + (-) my can eliminate x with 1-2 or 2-1 (m.n is a constant)
Constant: a rational number, such as 1.2.3, whose value does not change
Thank you. Please give me some points
Solving Fractional Inequalities
（x-4）/（X²+X-2）＞0
（x-4）/（X²+X-2）＞0
Two sets of inequalities are solved
【1】x-4＞0 x＞4
X²+X-2＞0 (x-1)(x+2)>0
X ＞ 1 and X ＞ - 2 take x ＞ 1
Because x ＞ 4 and X ＞ 1, take x ＞ 4
　　　X-1
（x-4）/（X²+X-2）＞0
(x-4)/(x-1)(x+2)>0
x> 4 or - 20
x> 4 or - 2
How to calculate X & # 178; + 3x-2 = 0
x²+3x=2
x²+3x+9/4=2+9/4
(x+3/2)²=17/4
x+3/2=±√17/2
x=(-3-√17)/2,x=(-3+√17)/2
One variable equation, one variable equation, two variable equation
One variable one time: solving equation (process formulation): (1) 4x-2 = 5 + 2x (2) 2 (X-5) + 2 = 3-4 (x-1) (3) x of three = 2x-1 (4) 2m-1 of six - 3m-1 of eight = 1. When x is the value, 3-2x of two and 2-x of three are opposite to each other?
Univariate quadratic: solve the following equation: (1) square of (x + 2) - 25 = square of 0 (2) x + 4x-5 = square of 0 (3) 2x - 7x + 3 = 0 (4) 7x (5x + 2) = 6 (5x + 2)
Bivariate linear: solve the following equations
(1) Y = x-3 and y-2x = 5 (2) 3m-2n = 5 and 4m + 2n = 9 (3) x-3y = 5 and 2x + y = 5 (4) 3x-5y = 7 and 4x + 2Y = 5
1 yuan per time:
(1)4x-2=5+2x
2x=7
x=3.5
(2）2（x-5）+2=3-4（x-1）
6x=15
x=2.5
(3) One third x = 2x-1
5x/3=1
x=3/5=0.6
(4) 6 / 2m-1 - 8 / 3m-1 = 1
8m-4-(9m-3)=24
m=-25
When what is the value of X, the formula 3 / 2-2x and 2 / 3-x are opposite to each other?
(3-2x)/2+(2-x)/3=0
9-6x+2-x=0
x=11/7
One variable quadratic: solve the following equation:
(1) The square of (x + 2) - 25 = 0
x+2=±5
X = 3 or - 7
(2) Square of X + 4x-5 = 0
(x+5)(x-1)=0
X = - 5 or 1
(3) The square of 2x - 7x + 3 = 0
(2x-1)(x-3)=0
X = 0.5 or 3
（4）7x（5x+2）=6（5x+2）
(7x-6)(5x+2)=0
X = 6 / 7 or - 2 / 5
Bivariate linear: solve the following equations
(1) Y = x-3 and y-2x = 5
(x-3)-2x=5
x=-8
y=-11
(2) 3m-2n = 5 and 4m + 2n = 9
The sum of the two formulas is 7m = 14, M = 2
n=0.5
(3) X-3y = 5 and 2x + y = 5
2(3y+5)+y=5
y=-5/7
x=20/7
(4) 3x-5y = 7 and 4x + 2Y = 5
Multiply the first by 2 and the second by 5
26x=39
x=3/2
y=-1/2
If you think the explanation is not clear enough, please ask
I wish: progress in learning!
One variable one time: solve the equation (process formulation)
（1）4x-2=5+2x
The result is that 2x = 7
The solution is x = 7 / 2
（2）2（x-5）+2=3-4（x-1）
It is reduced to 2x-8 = 7-4x
6X = 15
The solution is x = 5 / 2
(3) One third x = 2x-1
Multiply both sides of the equation by 3 to get x = 6x-3
The result is 5x = 3
The solution is x = 3 / 5
(4) unfold
One variable one time: solve the equation (process formulation)
（1）4x-2=5+2x
The result is that 2x = 7
The solution is x = 7 / 2
（2）2（x-5）+2=3-4（x-1）
It is reduced to 2x-8 = 7-4x
6X = 15
The solution is x = 5 / 2
(3) One third x = 2x-1
Multiply 6x by 3, and we get the equation x = 3
The result is 5x = 3
The solution is x = 3 / 5
(4) 6 / 2m-1 - 8 / 3m-1 = 1
If both sides of the equation are multiplied by 24, 4 (2m-1) - 3 (3m-1) = 24
The results are as follows: 1
The solution is m = - 25
When what is the value of X, the formula 3 / 2-2x and 2 / 3-x are opposite to each other?
From the question, 3 / 2-2x + 2 / 3-x = 0
3 (3-2x) + 2 (2-x) = 0 if both sides of the equation multiply by 6 at the same time
Simplify, 13-8x = 0
The solution is x = 13 / 8
One variable quadratic: solve the following equation:
(1) The square of (x + 2) - 25 = 0
The result is (x + 2) &# 178; = 25
The square root is x + 2 = ± 5
The solution is x = 3 or x = - 7
(2) Square of X + 4x-5 = 0
The factorization results in (x + 5) (x-1) = 0
The solution is x = 1 or x = - 5
(3) 2X = 7x + 3
The factorization results in (2x-1) (x-3) = 0
The solution is x = 1 / 2 or x = 3
（4）7x（5x+2）=6（5x+2）
The result is that (7x-6) (5x + 2) = 0
The solution is x = 6 / 7 or x = - 2 / 5
Bivariate linear: solve the following equations
（1）
y=x-3 ①
y-2x=5 ②
① Substituting in 2, we get
(x-3)-2x=5
The solution is x = - 8
Substituting the value of X into (1), we get
y=-8-3=-11
Therefore, the solution of the original equations is x = - 8, y = - 11
（2）
3m-2n=5 ①
4m+2n=9 ②
① + 2
7m=14
The solution is m = 2
Substituting M = 2 into 1, we get
6-2n=5
The solution is n = 1 / 2
Therefore, the solution of the original equations is m = 2, n = 1 / 2
（3）
x-3y=5 ①
2x+y=5 ②
② - 1 × 2
7y=-5
The solution is y = - 5 / 7
Substituting y = - 5 / 7 into (1), we get
x+15/7=5
The solution is x = 20 / 7
Therefore, the solution of the original equations is x = 20 / 7, y = - 5 / 7
（4）
3x-5y=7 ①
4x+2y=5 ②
① X 2 + 2 × 5
26x=39
The solution is x = 3 / 2
Substituting x = 3 / 2 into 2, we get
6+2y=5
The solution is y = - 1 / 2
Therefore, the solution of the original equations is x = 3 / 2, y = - 1 / 2
Grouping method, factoring problem,
y^4-4y^3+4y^2-1
Calculate to [y (x-1)] ^ 2-1
————
By the way, can the final result of factorization not contain brackets and constants?
y^4-4y^3+4y^2-1=（y^4-1）-4（y^3-y^2）=（y²+1）（y+1）（y-1）-4y²（y-1）=（y-1）【（y²+1）（y+1）-4y²】=（y-1）（y³-3y²+y+1）=（y-1）【（y³-1）-（3y²-y-2）】=（...
y^4-4y^3+4y^2-1
=(y²-2y)²-1
=(y²-2y+1)(y²-2y-1)
=(y-1)²(y²-2y-1)
Constant items in parentheses can have, usually without brackets.
[y(x-1)]^2-1
＝[y(x-1)＋1][y(x-1)－1]
＝（xy-y+1）（xy-xy-1）
The final result of factorization is factorization
y^4-4y^3+4y^2-1
=(y²-2y)²-1
=(y²-2y+1)(y²-2y-1)
=(y-1)²(y²-2y-1)
The final result of factorization does not contain brackets, and the constant term in brackets can exist
How to calculate X & # 178; - 3x + 2 = 0?
RT
2 is decomposed into - 1 × - 2, - 1 + (- 2) = - 3, and the formula is (x + a) (x + b) = x & # 178; + (a + b) x + ab
(X-2)(X-1)=0
X1=2 X2=1(X-2)(X-1)=0
How did you get here? Cross multiplication
It is also a method of factorization
There is a special section in the book of grade two
For example, the coefficient of quadratic term is 1 = 1 × 1, and the constant term is 2 = (- 2) × (- 1)
1　　　　　　－2
1　　　　　　－1
The sum of cross multiplication - 2 + (- 1) = - 3 is just equal to the coefficient of the first term
You can... Unfold
(X-2)(X-1)=0
X 1 = 2 x 2 = 1 question: (X-2) (x-1) = 0
How did you get here?
If you want to answer the equation of two variables only once
In ancient times, there is such a fable that donkeys and mules walk together. They carry different bags of goods. Each bag is the same weight. The donkey complains that the burden is too heavy. The mule says, "why do you complain? If you give me a bag, then my burden is twice as much as yours. If I give you a bag, then we just carry as much!" so what is the number of bags that the donkey carried?
Suppose that the number of bags the donkey was carrying was X
X+2+1=2（X-1）
X=5
Mules used to pack x + 2 = 7 bags
The donkey used to carry five bags of goods
Master to use group decomposition method to factorize~
x^4-x^2+4xy-4y^2
Factorization
Original formula = x ^ 4 - (x ^ 2-4xy + 4Y ^ 2)
=x^4-(x-2y)^2
=(x^2+x-2y)(x^2-x+2y)
=x^4 - (x-y)^2
=(x ^ 2 + X-Y) (x ^ 2-x + y) ask: sorry, wrong number, it should be y ^ 4-x ^ 2 + 4xy-4y ^ 2
Let P (x, y) be a circle (x-4) &# 178; + (Y-3) &# 178; = 4, find the maximum and minimum of T = y-3x, and find the maximum and minimum of Y / X
From t = y-3x, we get the equation of y = 3x + T into the circle, and get 10x & # 178; + (6t-26) x + (T & # 178; - 6T + 21) = 0. ① from P (x, y) on the circle, we know that equation 1 has a solution, and we get Δ≥ 0  - 2 √ 10-9 ≤ t ≤ 2 √ 10-9, so the maximum value of T = y-3x is 2 √ 10-9, and the minimum value is
When does a quadratic equation of one variable have two real roots?
If the quadratic equation of one variable has two real roots, it needs △ 0 (△ is a sign in Mathematics), △ B ^ 2-4ac (a is the coefficient of quadratic term, B is the coefficient of linear term, C is the number of constant term) for example: 4x ^ 2-8x + 12 = 0, then 4 is "a", - 8 is "B", 12 is "C" (an equation scribbled)