Good day, and welcome to Identity Boot Camp!

- Spoiler: No, this isn't going to be a page about finding yourself.

The term **"identities"** is used to mean several different things in mathematics. However, for the sake of today's problem and this introduction, we're going to define an identity as **"an equation that is true no matter what values you substitute in for the variables used in it."**

For example:

The Difference of Squares Identity

$(a+b)(a-b)=a^2-b^2$For any two real numbers $a$ and $b,$ the product of their sum and their difference is equal to the first of the two values squared minus the second of the two values squared.

Try testing any two real numbers into this equation for $a$ and $b,$ and you'll see that it *always* holds true *no matter what two numbers you choose.* For example, $(50+3)(50-3)=2500-9.$ You can use pen and paper or a calculator to check it if you want.

Identities can make solving a problem or simplifying out an algebraic description a lot easier, and today's challenge is no exception. In particular, the rest of this introduction defines and applies a set of relevant identities known as **the linearity of averages.**

Here's a problem to get things started. Try to solve it right now before we go through the solution analysis.

If, while solving this problem, you wrote out the statement "the average of $7x$ and $11x$ is $90$" algebraically, you probably realized that *a lot* of the algebra simplifies away: $\frac{7x + 11x}{2}=90$ simplifies to $\frac{18x}{2}=90,$ which again simplifies to $9x=90.$ Essentially, you just made use of this algebraic fact:

The average of $7a$ and $11a$ will always equal $9a,$ no matter what $a$ is. Written out algebraically, this corresponds to the identity $\frac{7a + 11a}{2} = 9a.$

Note that the equation $\frac{7a + 11a}{2} = 9a$ doesn't give you any new information about the variable $a$ because it's true **for all values of $a.$**

Additionally, this single claim about one average can be generalized into a much broader observation about averages that's important enough to have a name: "the linearity of averages."

The

linearity of averagesis a combination of these two identities:

If you multiply any two numbers $x$ and $y$ each by the same number, $a,$ then the average of the two new numbers is equal to $a$ times the average of the two original numbers:

$\frac{(a\times x)+(a\times y)}{2} = a \times \frac{x+y}{2}.$Additionally, if you add the same value, $k,$ to each of two numbers, then the average of the two new numbers will be increased by $k$ compared to the average of the two original numbers:$\frac{(x + k)+(y + k)}{2} = \frac{x+y}{2} + k.$

This also makes sense if you think about it on a number line. If you're counting out on a number line by $a\text{'s}, 9a$ will always be right in the middle between $7a$ and $11a.$

And this logic can be generalized to other numbers and larger averages. For example, the average of $2b, 17b,$ and $20b$ will be $\frac{2+17+20}{3} b = 13b.$

So, congratulations! You probably haven't found yourself yet, but you have at least managed to find a useful, mathematical truth.