**Q: Solve for x:**

**(a) x + 7 + 14x = 5(3x + 2) – 3 **

**(b) 2(x+5) + 10x = 12x – 15**

A: Okay, let’s start with the first one (a):

x + 7 + 14x = 5(3x + 2) – 3

Combine everything on the left side **and** multiply the 5 through to the (3x + 2) on the right side like so.. remember, the 5 multiples by both of the terms to get:

15x + 7 = 15x + 10 – 3

Now simplify the right side of the equation:

15x + 7 = 15x + 7

Now, subtract 15x from both sides:

15x + 7 – 15x = 15x + 7 – 15x

Notice the 15x on both sides go away and we get:

7 = 7

WHAT?? You are asking yourself: “Where the %@$! did the x go?!”

The x is gone. It turns out that x didn’t matter in this problem. 7 = 7 right? When does 7 = 7?? All the time! Therefore, x can be all real numbers… x can be anything and 7 will always equal 7…

So, the answer is:

x is all real numbers

OR, in interval notation:

(-∞, ∞)

Now for part (b):

2(x+5) + 10x = 12x – 15

Multiply the 2 through to the (x + 5) to get:

2x + 10 + 10x = 12x – 15

Combine the junk on the left:

12x + 10 = 12x – 15

Subtract 12x from both sides:

12x + 10 – 12x = 12x – 15 – 12x

10 = -15

WHAT??? Since when does 10 equal -15??? Never. Therefore, the answer is **no solutions**.