15 - x = 2 (x +3)

15y + 14 = 2(5y + 6)
1/2 (6x - 4) = 4x - 9
4(3d -2) = 8d - 5

please help me with these problems. I done other 20 problems just to let you know.

These are like the other problems I've done for you.

15 - x = 2 (x +3)
15 - x = 2x + 6
15 - 6 = 2x + x
9 = 3x
3 = x

Sure! I can help you with these algebraic problems. Let's go through each one step by step.

1) 15 - x = 2(x + 3)

To solve this equation, we need to isolate the variable on one side of the equation. Here's how to do it:

Expand the right side:
15 - x = 2x + 6

Combine like terms:
15 = 3x + 6

Subtract 6 from both sides:
15 - 6 = 3x
9 = 3x

Finally, divide both sides by 3 to solve for x:
x = 3

So the solution to the equation is x = 3.

2) 15y + 14 = 2(5y + 6)

First, simplify the right side by using the distributive property:
15y + 14 = 10y + 12

Next, subtract 10y from both sides to isolate the variable term:
15y - 10y + 14 = 10y - 10y + 12
5y + 14 = 12

Then, subtract 14 from both sides:
5y + 14 - 14 = 12 - 14
5y = -2

Finally, divide both sides by 5 to solve for y:
y = -2/5

3) 1/2(6x - 4) = 4x - 9

First, simplify the left side of the equation:
(6x - 4)/2 = 4x - 9

Next, divide both sides by 1/2 (or multiply by 2) to get rid of the fraction:
2(6x - 4)/2 = 2(4x - 9)
6x - 4 = 8x - 18

Now, subtract 6x from both sides to isolate the variable term:
6x - 6x - 4 = 8x - 6x - 18
-4 = 2x - 18

Lastly, add 18 to both sides to further isolate the variable term:
-4 + 18 = 2x - 18 + 18
14 = 2x

Divide both sides by 2 to find the value of x:
14/2 = 2x/2
7 = x

Therefore, the solution to the equation is x = 7.

4) 4(3d - 2) = 8d - 5

First, distribute the 4 to both terms inside the parentheses:
12d - 8 = 8d - 5

Next, subtract 8d from both sides to isolate the variable term:
12d - 8d - 8 = 8d - 8d - 5
4d - 8 = -5

Finally, add 8 to both sides to further isolate the variable term:
4d - 8 + 8 = -5 + 8
4d = 3

Divide both sides by 4 to solve for d:
d = 3/4

Therefore, the solution to the equation is d = 3/4.

I hope this helps! Let me know if you have any further questions.