If (A-2) x + (B + 1) y = 13 is a quadratic equation with respect to X and y, then a and B satisfy______ Conditions

If (A-2) x + (B + 1) y = 13 is a quadratic equation with respect to X and y, then a and B satisfy______ Conditions

∵ (A-2) x + (B + 1) y = 13 is a bivariate linear equation about X, y, ∵ a − 2 ≠ 0b + 1 ≠ 0, the solution is a ≠ 2, B ≠ - 1

Solving binary linear equations X: Y: z = 1:2:3, x + y + Z = 12

X：Y：Z=1：2：3
Then y = 2x, z = 3x
Substituting x + y + Z = 12, we get
X+2X+3X=12
The solution is x = 2,
Then y = 2x = 4
Z=3X=6

Given a binary linear equation system 40 (x + y) = 400, 200 (x − y) = 400, try to write an application problem according to the quantitative relationship of the given equation system and solve it

The distance between a and B is 400 meters. Snails a and B start from a and B respectively and meet in 40 minutes. If the two snails go in the same direction, it will take 200 minutes for a to catch up with B. what's the speed of snails a and B? Let the speed of snail A and snail B be x m / min and Y M / min respectively

Solving binary linear equations {x + y = 164.8x/60 + 12Y / 60 = 1.88

x+y=16
4.8x/60+12y/60=1.88
The following formula is simplified to 2x + 5Y = 47 (multiply both sides by 600 and divide by 24)
Multiply the first formula by 2 to get 2x + 2Y = 32
By subtracting and eliminating, 3Y = 15
So y = 5, x = 11

The difference between a number and its reciprocal is 14 and 14 / 15. What is the number?
/Yes?

15
15-1/15=14+14/15

2X + 2Y = y is a quadratic equation of two variables

Yes, but there are countless solutions to your problem. You need an equation to solve one

What is the existence theorem of function zeros?

If the image of function y = f (x) on interval [a, b] is an uninterrupted curve, and f (a) f (b)

It is known that the quadratic equation of one variable (2m-1) x'2-2 with sign (MX) + 1 = 0 has two unequal real roots (1) the value range of M; (2) when m + M 1 / 2 = 11
Finding the value of root sign M-1 / 2 of root sign M

(1)
∵ the univariate quadratic equation (2m-1) x'2-2 with sign (MX) + 1 = 0 has two unequal real roots
∴△=(-m√2)²-4(2m-1)＞0
2m²-8m+4＞0
m²-4m+4＞2
(m-2)²＞2
∴m-2＞2 m-2＜-2
Ψ m ＞ 4 or m ＜ 0
(2) When m + M 1 / 2 = 11
m+1/m=11
m-2+1/m=11-2
[√m-1/(√m)]²=9
∴√m-1/(√m)=±3

It is known that the quadratic function y = x2 + (2m-1) x + M2 + 2 has the minimum value. 2. (1) find the value of M; (2) find the vertex coordinates and symmetry axis of the quadratic function

(1) According to the meaning of the title, we get 4 (M2 + 2) − (2m − 1) 24 = 2, and the solution is m = 14; (2) when m = 14, y = x2-12x + 3316, the symmetry axis line x = - − 122 = 14, and the vertex coordinates are (14, 2)

How to solve two equations with different unknowns? 3 / 4x-1.4 = 3.6 = 2 / 3x

3/4x-1.4=3.6+2/3x
3/4x-2/3x=1.4+3.6
x/12=5
x=5*12
x=60