The following equations are solved by substitution elimination method: 1. X + y = 11, X-Y = 7, 2. 3x-2y = 9, x + 2Y = 3 x+y=11 x-y=7 3x-2y=9 x+2y=3

The following equations are solved by substitution elimination method: 1. X + y = 11, X-Y = 7, 2. 3x-2y = 9, x + 2Y = 3 x+y=11 x-y=7 3x-2y=9 x+2y=3

Substitution elimination method: change the coefficient of an unknown number of one equation into 1, and substitute it into another equation. For example: x + y = 11, ① X-Y = 7, ② from ① get: y = 11-x, ③ substitute ③ into ② get: X - (11-x) = 7, x = 9. The solution of the system of equations is {x = 9 {y = 2 {3x-2y = 9}

The following equations, x-2y = 3, 3x + y = 2, are solved by substitution elimination method

1、 By substituting x-2y = 3: x = 3 + 2Y into the second formula: 3 (3 + 2Y) + y = 27y = - 7, y = - 1, then x = 1; by substituting 3x + y = 2: y = 2-3x into the first formula: X-2 (2-3x) = 37x = 7, then y = - 1

Ternary linear equations 3x + y + 2Z = 11 2x + y + 3Z = 12 x-y-4z = - 9

x=2,y=2,z=3

Review of mathematical Algebra (5 20:14:3)
1 it is known that when x = 1, the value of the 5th power of the algebraic formula 2x - BX to the 3rd power + cx-4 is 9, then when x = - 1, the value of the 5th power of the algebraic formula ax - BX to the 3rd power + cx-4 is (& # 160; & # 160; & # 160; & # 160; & # 160; & # 160;)
2 given that a = 2 / 5, B = 6 / 5, find the value of the second power-3 (2a + b) of the algebraic formula 2 (2a + b)
3 has two polynomials: a = 2A to the second power - 4A + 1, B = 2 (a to the second power - 2A) + 3. When a takes any rational number, please compare the size of a and B

1 it is known that when x = 1, the value of the 5th power of the algebraic formula 2x-bx + cx-4 is 9, then when x = - 1, the value of the 5th power of the algebraic formula ax-bx + cx-4 is ()
The value of the algebraic formula a times the 5th power of X - B times the 3rd power of X + cx-4 is 9
Substituting x = 1 into A-B + C = 13
Substituting x = - 1 gives - A + b-c-4 = - (a-b + C) - 4 = - 13-4 = - 17
Please write the title clearly next time
2 given that a = 2 / 5, B = 6 / 5, find the value of the second power-3 (2a + b) of the algebraic formula 2 (2a + b)
This problem directly into the data to solve the result of 2 is not difficult, right
3 has two polynomials: a = 2A to the second power - 4A + 1, B = 2 (a to the second power - 2A) + 3. When a takes any rational number, please compare the size of a and B
If a takes any rational number, it holds. If a = 1, then b > a

Find the value of the algebraic formula (2a2-5a) - 2 (3a + 5-2a2), where a = - 1
First simplify and then evaluate: the square of 4 (a + b) - the square of 7 (a + b) + (a + b), where a = 1 / 2, B = 1 / 3
First simplify, then evaluate: - 2 (square of mn-3m) - [square of m-5 (square of mn-m) + 2Mn], where M = 1, n = - 2

The value of the algebraic formula (2a2-5a) - 2 (3a + 5-2a2), where a = - 1. = 2A ^ 2-5a-6a-10 + 4A ^ 2 = 6A ^ 2-11a-10 = 6 * (- 1) ^ 2-11 * (- 1) - 10 = 6 + 11-10 = 7, first simplify and then evaluate: the square of 4 (a + b) - the square of 7 (a + b) + (a + b), where a = 1 / 2, B = 14 / 3 (a

If the value of formula 2A2 + 3A + 1 is 5, then the value of formula 4a2 + 6A + 8 is 5

From 2A & # 178; + 3A + 1 = 5, we get: 2A & # 178; + 3A = 4
4a²+6a+8=2(2a²+3a)+8=8+8=16

Given the algebraic formula AX2 + BX + C, when x = 0, its value is - 3; when x = - 3, its value is 0; when x = 2, its value is 5,
(2) Find the value of algebra when x = - 0.5

Substitute x = 0 into ax ^ 2 + BX + C = - 3, 0 + 0 + C = - 3
c=-3
Substitute x = - 3, C = - 3 into ax ^ 2 + BX + C = 0, 9a-3b = 3
Substitute x = 2, C = - 3 into ax ^ 2 + BX + C = 5.4A + 2B = 8
LIANLI 9a-3b = 3
4a+2b=8
The solution is a = 1
b=2
(2) Find the value of algebra when x = - 0.5
x^2+2x-3=0
When x = - 0.5, - 3.75

Given the algebraic formula AX2 + BX + C, when x = - 1, its value is 4; when x = 1, its value is 0; when x = 2, its value is 7, then when x = - 2, its value is 0
Good answer,

From the meaning of the title: A-B + C = 4
a+b+c=0
4a+2b+c=7
The solution is a = 3
b=-2
c=-1
So the algebraic formula is 3x ^ 2-2x-1
So when x = - 2, the value is 3 * 4 + 4-1 = 15

X = 3, the algebraic formula AX2 + BX + 8 to is 12. Find the value of x = 3, the algebraic formula AX2 + bx-5
Solve by equation
It should be: x = 3, the cube of the algebraic formula ax + BX + 8 to 12. Find the value of x = 3, the cube of the algebraic formula ax + bx-5

Substituting x = 3 into the algebraic expression AX2 + BX + 8, we get 9A + 3B + 8 = 12
So 9A + 3B = 4
Then we substitute x = 3 into the formula AX2 + bx-5, and get 9A + 3b-5 = 4-5 = - 1
So, when x = 3, the value of the algebraic expression AX2 + bx-5 is - 1
The question has been changed - the answer is the same!
The only way to do that is to change 9 to 27

Given the algebraic formula AX2 = BX +, when x = - 1, its value is 4; when x = 1, its value is 8; when x = 2, its value is 25. When x = 3, its value is 4
Square of a + C

The algebraic formula should be ax ^ 2 + BX + C. if so, the solution is as follows:
A-B + C = 4 (1)
a+b+c=8 (2)
4a+2b+c=25 (3)
(2) After - (1), B = 2
After substituting B = 6 into (2) and (3), (3) - (2), a = 5 is obtained
When a = 5 and B = 2 are substituted into (2), C = 1 is obtained, then when x = 0
ax^2+bx+c=5*3^2+2*3+1=52