Question

-2х-0,8у-1=0. -2х+0,2у=0

Answer 1

МЕТОДОМ СЛОЖЕНИЯ
-2х-0.8у-1=0 /*(-1)
-2х+0.2у=0


2х+0.8у+1=0
+
-2х+0.2у=0


у+1=0
-2х+0.2у=0


у=-1
-2х-0.2=0


у=-1
-2х=0.2


у=-1
х=-0.1

(-0.1; -1)

Answer 2

Я так понимаю,что это система ,тогда
1)-2х=1+0.8у
2)1+0.8у+0.2у=0
у=-1
3)-2х=0.2
х=-0.1

Answer 3

It looks like you have a system of two equations with two variables:

Equation 1: -2x - 0.8y - 1 = 0 Equation 2: -2x + 0.2y = 0

You can solve this system of equations to find the values of x and y that satisfy both equations. There are various methods to solve systems of equations, such as substitution, elimination, or matrix methods. Let's use the elimination method to solve this system.

First, let's multiply the second equation by 4 to make the coefficients of x in both equations the same:

2 * (-2x + 0.2y) = 0 -4x + 0.8y = 0

Now we have two equations with the same coefficient for x:

Equation 1: -2x - 0.8y - 1 = 0 Equation 2 (multiplied by 2): -4x + 0.8y = 0

Now, add Equation 1 and Equation 2 to eliminate y:

(-2x - 0.8y - 1) + (-4x + 0.8y) = 0 -6x - 1 = 0 -6x = 1 x = -1/6

Now that we have the value of x, we can substitute it back into Equation 2 to solve for y:

-2x + 0.2y = 0 -2 * (-1/6) + 0.2y = 0 1/3 + 0.2y = 0 0.2y = -1/3 y = -1.5

So, the solution to the system of equations is x = -1/6 and y = -1.5.