If x = 1, y = - 1 and x = 3, y = 5 are solutions of the equation AX + by + 2 = 0, try to judge whether x = 4, y = 8 are solutions of the equation AX + by + 2 = 0?

If x = 1, y = - 1 and x = 3, y = 5 are solutions of the equation AX + by + 2 = 0, try to judge whether x = 4, y = 8 are solutions of the equation AX + by + 2 = 0?

Substituting x = 1, y = - 1 and x = 3, y = 5 into ax + by + 2 = 0 respectively, we get the following results:
a-b+2=0
3a+5b+2=0
The solution is as follows
a=-3/2
b=1/2
So the original equation is:
-3/2x+1/2y+2=0
Substituting x = 4, y = 8, we get the following result
-3/2×4+1/2×8=-6+4=-2≠0
So x = 4 and y = 8 are not solutions of the equation AX + by + 2 = 0

If x = 1, y = - 1 and x = 3, y = - 5 are solutions of the equation AX + by + 2 = 0, try to judge whether x = - 4, y = 9 are solutions of the equation AX + by + 2 = 0?

I'm glad to answer your question. Take x = 1, y = - 1 and x = 3, y = - 5 into ax + by + 2 = 0 to get the equation system a - B + 2 = 0. ① 3a-5b + 2 = 0. ② take ① * 3 to get 3a-3b + 6 = 0. ① - ② to get 2B + 4 = 0. To get b = - 2. ③ to get a - (- 2) + 2 = 0A + 2 + 2 = 0. To get a = - 4. Then the equation is - 4x-2y + 2 = 0

It is known that {x = 1, y = 3 and {x = 3, y = 5 are two sets of solutions of the equation AX + by = 30. Find the values of a and B

The {x = 1, y = 3 and {x = 3, y = 5 are two sets of solutions of the equation AX + by = 30
The results show that {a + 3B = 30
3a+5b=30
The solution is {a = - 15, B = 15

If {x = 1, y = - 1 {x = 2, y = 2 and {x = 3, y = C are solutions of the equation AX + by + 2, find the value of C

X = 1, y = - 1 substituting ax + by + 2 = 0
A-B = - 21
Substituting x = 2, y = 2 into ax + by + 2 = 0
2a+2b+2=0
A + B = - 1,2
Substituting x = 3, y = C into ax + by + 2 = 0
3A + BC = - 2, 3
Formula 1: a = B-2 is substituted into formula 3
3(b-2)+bc =-2
3b-6+bc=-2
3b+bc=4
Formula B (3 + C) = 4
Formula 1 - Formula 2: - 2b = 1
b= -1/2
Substituting B = - 1 / 2 into formula 4
-1/2(3+c)=4
-(3+c)=8
-3-c =8
-c =11
c=11

X-24 = 127, 36-4x = 4

Child, what grade are you in? You have to do this by yourself. Forget it. I'll help you
(1) x-24=127
x=151
(2) 36-4x=4
36-4=4x
4x=32
x=8

1.99 × 4.3 10.1 × 7.6 simple operation

=(2 minus 0.1) x4.3
=2x4.3 minus 0.1x4.3
=8.6 minus 0.43
=8.17

Volume one (urgent, urgent.)

s＝ah÷2 ＝16*9.5÷2 ＝152÷2 ＝76m2

After two consecutive price cuts of 20%, the price is 144 yuan. How much is the original price?

What is the original price
144÷（1-20%）÷（1-20%）
=144÷0.8÷0.8
=225 yuan, just one step to give a good comment!

Given the function f (x) = the square of X + ax + 3, the domain of definition is r
(1) If f (2-x) = f (2 + x), find the value range of real number a?
(2) When x belongs to [- 2,4], find the maximum value of function f (x)?
(3) When x belongs to [- 2,2], it is true that f (x) is greater than or equal to a?
I beg the elder brothers and sisters to help my younger brother. I'm broken in thinking,

f(x)=x^2+ax+3=(x+a/2)^2+3-a^2/4
1) It shows that x = 2 is the axis of symmetry, so 2 = - A / 2, a = - 4
2) Because the opening is upward, the maximum value is obtained at the end point, and the maximum value is obtained at the end point far from the axis of symmetry
If - A / 2 = - 2, the maximum is f (4) = 19 + 4a
If - A / 2 > 1, that is a

Mathematical problems on quadratic equation of one variable
For any real number k, the equation (K & sup2; + 1) x & sup2; - 2 (a + k) & sup2; X + K & sup2; + 4K + B = 0 always has a root of 1
(1) find real numbers a and B (2) find the range of another root

1. X = 1, bring in (4-4a) k-2a ^ 2 + 1 + B = 0, because for any k, there must be 4-4a = 0, a = 1, so B = 1
2. According to the relationship between root and coefficient, the product of two is (k ^ 2 + 4K + 1) / (k ^ 2 + 1). Because one root is 1, the other follows (k ^ 2 + 4K + 1) / (k ^ 2 + 1), that is, 1 + 4K / (k ^ 2 + 1). When k = 0, the other root is 1. When k is not 0, look at the part of 4K / (k ^ 2 + 1), divide it by K, that is, 4 / (K + 1 / k), and then divide it into two cases where k is greater than 0 and less than 0, So the range of 4 / (K + 1 / k) is [- 2,0) and (0,2], and then synthesize the case of k = 0, so another following range is [- 1,3]