# Question #24e48

##### 1 Answer

#### Answer:

Your equation has an infinite number of solutions.

#### Explanation:

Your starting expression looks like this

#6 * (2x + 4) = 10x + 24 + 2x#

Notice that your equation contains *two types* of terms

*terms that contain*#x# *terms that do not contain*#x#

Your goal here is to get all the tems that contain

So, start by expanding the paranthesis by multiplying both terms by

#6 * (2x + 4) = 6 * 2x + 6 * 4 = 12x + 24#

The equation now looks like this

#12x + 24 = 10x + 24 + 2x#

Notice that you can group two terms that contain

#12x + 24 = underbrace(12x)_(color(blue)(10x + 2x)) + 24#

Notice that we are left with *the same expression* on both sides of the equation

#12x + 24 = 12x + 24#

In this case, you would say that the equation has an **infinite number of solutions** because you can plug in any value of

More specifically, this is reduced to

#color(red)(cancel(color(black)(12x))) - color(red)(cancel(color(black)(12x))) = color(red)(cancel(color(black)(24))) - color(red)(cancel(color(black)(24)))#

#0 = 0#

This is true regardless of the value of *infinite number of solutions*.