Question #24e48

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 #x# on one side of the equation, and all the terms that do not contain #x# on the othe side of the equation.

So, start by expanding the paranthesis by multiplying both terms by #6#

#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 #x# on the right-hand side of the equation to get

#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 #x# and the equation will be true!

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 #x#, which is why the equation is said to have an infinite number of solutions.