# Venn diagram question

Here is my question.

A math examination has three questions. Twenty-six students took the examination, and every student answered at least one question. Six students did not answer the first question; twelve did not answer the second question; and five did not answer the third question. If eight students answered all three questions, how many students answered exactly one question?

The answer and the Venn diagram for the exercise is:

From the regions in the Venn diagram, we have

$a + b + c + d + e + f + g = 26$ : the total number of students

$g = 8$ : all three questions answered

$b + f + c = 6$ : Question 1 is not answered

$a + e + c = 12$ : Question 2 is not answered

$a + d + b = 5$ : Question 3 is not answered

My confusion starts here: Why do I want to find $a + b + c$, as stated below?

We want to find $a + b + c$, and we know that $a + b + c > 3$. Adding the last three equations, we have

Why do I do this next step? $2(a + b + c) + e + d + f = 23$

and

Why do I do this next step? $a + b + c + 23 - 2(a + b + c) + 8 = 26$:

How do I get this next step? Why is a + b + c = 5? So we conclude that $a + b + c = 5$.

Thanks for any help you can provide. I have been at this section on Venn diagrams all week and I just can't seem to get it.

We also know $g=8$, or equivalently $$a+b+c+d+e+f=26-8=18.$$
By summing $$\color{blue}{b}+f+\color{red}{c}=6,$$ $$\color{green}{a}+e+\color{red}{c}=12,$$ $$\color{green}{a}+d+\color{blue}{b}=5$$ we're essentially double-counting $a$, $b$ and $c$ (those who answered one question) whereas we're single-counting $d$, $e$ and $f$ (those who answered two questions). This imbalance between $a+b+c$ and $d+e+f$ can be exploited: it enables us to separate $a+b+c$ out from the first equation.
The rest is just arithmetic. Let $X=a+b+c$, then the final equation implies $X+23-2X+8=26$, and we solve for $X$.