Question

1.²/9x=1,82.-4x=23.9x-8=4x+124.9-7(x+3)=5-6x​

Answer 1

1) 2/9х=18

9х=9

х=1

....................

2)-4х=2

х=2/-4

х=-0,5

..............

3)9х-8=4х+12

9х-8-4х-12=0

5х=-20

х=-4

.................

4)9-7(х+3)=5-6х

9-7х-21-5+6х=0

-х=-16 (все уравнение делим на (-1))

х=16

Answer 2

To solve each of these equations, we'll isolate the variable (x) on one side of the equation. Let's go through them one by one:

1. 2/9x = 1

To solve for x, we need to get rid of the fraction on the left side. We can do this by multiplying both sides of the equation by 9/2:

(9/2) * (2/9x) = (9/2) * 1

The 2/9 on the left side cancels out, leaving us with:

x = 9/2

So the solution to this equation is x = 4.5.

2. -4x = 23

To solve for x, divide both sides of the equation by -4:

(-4x) / -4 = 23 / -4

The -4 on the left side cancels out, and we get:

x = -23/4

So the solution to this equation is x = -5.75.

3. 9x - 8 = 4x + 12

To solve for x, we need to isolate the x term on one side and the constants on the other side. Let's move the 4x term to the left side by subtracting 4x from both sides:

9x - 4x - 8 = 4x - 4x + 12

Simplifying:

5x - 8 = 12

Next, let's move the constant term to the right side by adding 8 to both sides:

5x - 8 + 8 = 12 + 8

Simplifying:

5x = 20

Finally, to find the value of x, divide both sides of the equation by 5:

(5x) / 5 = 20 / 5

x = 4

So the solution to this equation is x = 4.

4. 9 - 7(x + 3) = 5 - 6x

Let's start by distributing the -7 on the left side:

9 - 7x - 7(3) = 5 - 6x

Simplify:

9 - 7x - 21 = 5 - 6x

Combine like terms:

-7x - 12 = 5 - 6x

Now, let's get all the x terms on one side by adding 6x to both sides:

-7x + 6x - 12 = 5 - 6x + 6x

Simplify:

-x - 12 = 5

Next, let's move the constant term to the right side by adding 12 to both sides:

-x - 12 + 12 = 5 + 12

Simplify:

-x = 17

Finally, to find the value of x, multiply both sides of the equation by -1:

(-1) * (-x) = (-1) * 17

x = -17

So the solution to this equation is x = -17.

Remember to always check your solutions by plugging them back into the original equations to ensure they are valid.