Equation 3x + 2 = 2 + 5x

Equation 3x + 2 = 2 + 5x

3x＋2＝2＋5x
2-2=5x-3x
0=2x
x=0

How to solve the equation (5x + 1): (3x-1) = 9:5

9(3x+1)=5(5x+1)
27x-9=25x+5
2x=14
x=7

5x-2.6 + 0.4 = 3x + 1.9 use equation solution, urgent

5X-2.6+0.4=3X+1.9
5x-3x=1.9+2.6-0.4
2x=4.1
x=2.05

Find the locus of the point with the same distance between two straight lines 3x + 4y-12 = 0 and X + Y-1 = 0

This trajectory is the angular bisector of two straight lines
If the point is (x, y), then the distance between the point and the two straight lines is equal, so the formula of the distance between the point and the straight line is used
|3x+4y-12|/√5＝｜x+y-1｜/√2
This is two linear equations, and over
The intersection of 3x + 4y-12 = 0 and X + Y-1 = 0 (- 8,9)

The distance between two parallel lines 3x + 4Y + 7 = 0 and 3x + 4Y + 17 = 0 is

Take a point on 3x + 4Y + 7 = 0, such as (- 1, - 1)
The distance from him to a straight line is what he wants
So the distance = | - 3-4 + 17 | / √ (3 & # 178; + 4 & # 178;) = 10 / 5 = 2

What is the minimum distance between the point on the circle x ^ 2 + y ^ 2 = 1 and the line 4x-3y + 25 = 0

Calculate the distance from the center of a circle to a straight line with the formula of distance from a point to a straight line. If it is greater than the radius, then reduce the radius. If it is less than the radius, then it is 0
d=25/√25=5>1
So the minimum distance is 5-1 = 4

The ratio of the distance between the point P and the vertex f (2,0) and the distance of its straight line x = 8 is 1:2
The more urgent, the better

Let P coordinate (x, y) pf ^ 2 = (X-2) ^ 2 + y ^ 2p be the distance from the line x = 8 d = | X-8 | so PF: D = 1:2pf ^ 2: D ^ 2 = 1:4 (X-8) ^ 2 = 4 [(X-2) ^ 2 + y ^ 2] x ^ 2-16x + 64 = 4x ^ 2-16x + 16 + 4Y ^ 23x ^ 2 + 4Y ^ 2 = 48, that is, the trajectory equation is: 3x ^ 2 + 4Y ^ 2 = 48, that is, ellipse x ^ 2 / 16 + y ^ 2 / 12 = 1

The ratio of the distance from the moving point P (x, y) to the fixed point F (1, 0) to the distance from it to the fixed line x = 4 is 1:2
The ratio of the distance from the moving point P (x, y) to the fixed point F (1, 0) to the distance from it to the fixed line x = 4 is 1:2

4[(x-1)2+y2]=(x-4)2
3x2+4y2-12=0

The trajectory equation of a point in the plane whose distance to the vertex m (2.2) is equal to that to the fixed line x + y-4 = 0 is ()

The trajectory of such a point is: parabola

Find the distance between the following two parallel lines. 3x plus 4Y minus 12 equals zero and 6x plus 8y plus 11 equals 0

3x plus 4Y minus 12 equals zero and 6x plus 8y plus 11 equals 0
The two lines are
6x+8y-24=0
6x+8y+11=0
therefore
Distance = | 11 - (- 24) | / √ (6 square + 8 square)
=35÷10
=7/2