Simplify or evaluate - 3 (5x-7y) + 6 (2x-4y) and a - (2a-7b) + 3 (a-2b), where a = - 2 and B = - 100

Simplify or evaluate - 3 (5x-7y) + 6 (2x-4y) and a - (2a-7b) + 3 (a-2b), where a = - 2 and B = - 100

-3 (5x-7y) + 6 (2x-4y) and a - (2a-7b) + 3 (a-2b), where a = - 2 and B = - 100
=-15x+21y+12x-24y =a-2a+7b+3a-6b
=-3x-3y =2a+b
a=-2,b=-100
2a+b=2*(-2)+(-100)=-4-100=-104
Original formula = - 15x + 21y + 12x-24y = - 3x + (- 3) y = - 3 (x + y) original formula = a-2a + 7b + 3a-6b = 2A + B = 2 * (- 2) - 100 = - 104
The original formula = a-2a + 7b + 3a-6b = 2A + B = 2 * (- 2) - 100 = - 104
Given (5x-2y-3) &# 178; + 2x-3y + 1 = 0, find the value of X + y, give me a way to solve this problem
The sum of squares and absolute values are nonnegative. The sum of two numbers is 0, which means that two numbers can only be 0,
therefore
5x-2y-3=0
2x-3y+1=0
The results of solving the equations are: x = y = 1
So x + y = 2
2x-3y + Z + (x + 2y-z) & # 178; = 0 and X + y + Z = 11 form a system of linear equations with three variables, and find x, y, Z
From 2x-3y + Z + (x + 2y-z) ² = 0
We obtain 2x-3y + Z = 0
x+2y+z=0 ②
And X + y + Z = 11
②-③ :y=-11 ,x+z=22
Substituting ①: 2x + Z = - 33
∴x=-55 ,z=77
∴{x=-55 ,y=-11 ,z=77
Given 2x + Y / 2 = 5x + 2Y / 4 = 1, find the value of X + 2Y + 1 / 2x-3y + 7
Given that the solutions of {2x + 5Y = - 6, ax-by + - 4 and {3x-5y = 16, 6x + ay = - 8 are the same, find the value of (a + b) ^ 2
Party A and Party B solve the equations {ax + 5Y = 15,1.4x-by = 5,2 together. Because party a misinterprets a in equation 1, the end {x = - 3, y = 1 is obtained, while Party B misinterprets B in equation 2, the solution is {x = 5, y = 4. Try to find the value of a ^ 2006 + (- 1 / 10B) ^ 2005
emergency
competent
1. It should be [2x + y] / 2 = [5x + 2Y] / 4 = 1, find the value of [x + 2Y + 1] / [2x-3y + 7] = 4x + 2Y = 4,5x + 2Y = 4x = 0, y = 1 [x + 2Y + 1] / [2x-3y + 7] = 12.2x + 5Y = - 6,3x-5y = 16, x = 2, y = - 2A = 10, B = - 12 (a + b) ^ 2 = 43
1. Solve the equations 2x + Y / 2 = 1,5x + 2Y / 4 = 1, and then substitute them into the required formula
5/4
2. Because the two equations have the same solution, we can solve the equations 2x + 5Y = - 6, 3x-5y = 16. We can get x = 2, y = - 2, and substitute the original formula to continue to solve the equations to get ab
Four
3. Because a misinterprets a in equation 1, B should be able to get the correct one. Take x = - 3, y = 1 generation, 1.4x-by = 5 to get B. similarly, expand x = 5, y = 4
1. Solve the equations 2x + Y / 2 = 1,5x + 2Y / 4 = 1, and then substitute them into the required formula
5/4
2. Because the two equations have the same solution, we can solve the equations 2x + 5Y = - 6, 3x-5y = 16. We can get x = 2, y = - 2, and substitute the original formula to continue to solve the equations to get ab
Four
3. Because a misread a in equation 1, B should be able to get the correct one. Take x = - 3, y = 1 generation, 1.4x-by = 5 to get B. similarly, take x = 5, y = 4, ax + 5Y = 15 to get a
The distance between parallel lines 3x + 4y-9 = 0 and 6x + 8y + 2 = 0 is
3x + 4y-9 = 0 and 6x + 8y + 2 = 0
Namely
6X + 8y-18 = 0 and 6x + 8y + 2 = 0
Using the formula of distance between parallel lines
d=|-18-2|/√(3²+4²)=20/5=4
Known 2x + 4Y = 1088, 3x + 2Y = 1264, find the value of X, y, find the process, thank you!
Equation two times 2-equation one
x=360
Substituting it into the equation
y=92
Equation 2 times 2 minus equation 1 gives 4x = 1240, x = 310, and substituting equation 1 gives y = 117
The quadratic function f (x) = AX2 + BX (a, B are constants, and a ≠ 0) satisfies the following conditions: F (2) = 2, and the equation f (x) = x has equal roots
The equation f (x) = x has equal roots, so ax ^ 2 + (B-1) x = 0 has equal roots
So (B-1) ^ 2-4a = 0 (1)
And f (2) = 2, so
4a+2b=2 (2)
From (1) (2), we get (B-1) ^ 2 + 2b-2 = 0
That is, (B-1) (B-1 + 2) = 0
B = 1 or - 1
A = 0 or 1
Because a is not equal to 0
So a = 1
b=-1
If integers x and y satisfy the inequality x2 + Y2 + 1 ≤ 2x + 2Y, then the value of X + y has ()
A. 1 B. 2 C. 3 d. 4
The deformation of x2 + Y2 + 1 ≤ 2x + 2Y is that, x2-2x + 1 + y2-2y + 1 ≤ 1, (x-1) 2 + (Y-1) 2 ≤ 1, and (x-1) 2 ≥ 0, (Y-1) 2 ≥ 0, the following conditions can be obtained: X − 1 = 0y − 1 = 0 or X − 1 = ± 1y − 1 = 0 or X − 1 = 0y − 1 = ± 1, so the value of X + y has 2, 3 and 1, so it should be selected: C
If the set a = {x | x2 + (K-3) x + K + 5 = 0, X ∈ r}, a ∩ R + ≠ Φ, then the value range of real number k is______ .
From the meaning of the question, we know that the equation x2 + (K-3) x + K + 5 = 0 must have a positive root. From x2 + (K-3) x + K + 5 = 0, we get - k = x2 − 3x + 5x + 1 = & nbsp; X + 1 + 9x + 1 − 5 ≥ 1, so fill in (- ∞, - 1)]
3x+2y+z=9 x-3y-2z=1 -x+4y=5
3x+2y+z=9 (1)
x-3y-2z=1 (2)
-x+4y=5 (3)
2(1)+(2)
7x+y = 19 (4)
4(4) -(3)
29x=71
x=71/29
from (3)
y = 54/29
from (1)
3x+2y+z=9
3(71/29)+2(54/29)+z=9
z =-60/29
From Formula 1, we can get: 6x + 4Y + 2Z = 18
1 + 2 7x + y = 19
Then add x = 4y-5 to get y = 54 / 29, x = 71 / 29. Z = - 60 / 29