Question

-5 (2-2х)=2(х-3)+42х-4(х+7)=3(2-х)​

Answer 1

Ответ:

1) - 5(2-2х)=2(х-3)+4

-10+10х=2х-6+4

10х-2х=4-6+10

8х=8

Х=1

2)2х-4(х+7)=3(2-х)

2х-4х-28=6-3х

2х-4х+3х=6+28

Х=34

Answer 2

To solve this system of equations, we'll simplify and solve each equation one at a time. Let's start with the first equation:

1. -5(2 - 2x) = 2(x - 3)

First, distribute the constants on both sides:

-10 + 10x = 2x - 6

Now, move the terms involving x to one side of the equation and the constants to the other side:

10x - 2x = -6 + 10

Combine like terms:

8x = 4

Now, divide both sides by 8 to isolate x:

x = 4/8

x = 1/2

Now that we've found the value of x, we can substitute it into the second equation and solve for x:

2. 42x - 4(x + 7) = 3(2 - x)

First, distribute the constants on both sides:

42x - 4x - 28 = 6 - 3x

Now, move the terms involving x to one side of the equation and the constants to the other side:

42x - 4x + 3x = 6 + 28

Combine like terms:

41x = 34

Now, divide both sides by 41 to isolate x:

x = 34/41

So, the solution to the system of equations is:

x = 1/2 x = 34/41