# Question #24e48

Oct 12, 2015

Your equation has an infinite number of solutions.

#### Explanation:

Your starting expression looks like this

$6 \cdot \left(2 x + 4\right) = 10 x + 24 + 2 x$

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 $6$

$6 \cdot \left(2 x + 4\right) = 6 \cdot 2 x + 6 \cdot 4 = 12 x + 24$

The equation now looks like this

$12 x + 24 = 10 x + 24 + 2 x$

Notice that you can group two terms that contain $x$ on the right-hand side of the equation to get

$12 x + 24 = {\underbrace{12 x}}_{\textcolor{b l u e}{10 x + 2 x}} + 24$

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

$12 x + 24 = 12 x + 24$

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

$\textcolor{red}{\cancel{\textcolor{b l a c k}{12 x}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{12 x}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{24}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{24}}}$

$0 = 0$

This is true regardless of the value of $x$, which is why the equation is said to have an infinite number of solutions.