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# Algebra: Solving for x

Q:  Solve for x:

(a)  x + 7 + 14x = 5(3x + 2) – 3

(b)  2(x+5) + 10x = 12x – 15

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…

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.