Why is this system of equations unsolvable? The system is:
\begin{cases}
3a + 5b = 2\\
15a + 15b = ?
\end{cases}
Can't we just do:
\begin{gather}
3a = 2 - 5b\\[2ex]
a = \frac{2 - 5b}{3}
\end{gather}
Then we plug in $a$ in terms of $b$ into the second equation, which gives:
\begin{gather}
\frac{15 (2 - 5b)}{3} + 15b = ?\\
10 - 25b + 15b = ?\\
-10b = -10\\
b = 1
\end{gather}
Then we plug in $b$ into the first equation, to get:
\begin{gather}
3a + 5 = 2\\
3a = -3\\
a = -1
\end{gather}
We've solved for both $a$ and $b$ and we see that $15a + 15b$ is $0$, therefore $? = 0$.
 A: Going from line 9 to line 10 you assumed that $?=0$, so it should be no surprise that you ended up with $?=0$.
I'm really not sure what the question means but perhaps it means "if $3a+5b=2$, can you find the value of $15a+15b$?"
If this is the question then the answer is "no you can't" because $?$ could actually be any real number.  For example, if $a=1$ and $b=-\frac{1}{5}$, then $?=12$.
A: You have three unknowns and two equations. It should be immediately obvious that this isn't solvable.
As David accurately pointed out, you accidently set ? = 0 in your work.
A: Up until here it's correct:
$$
10 - 25b + 15b = ?
$$
The problem is the next step:
$$
-10b = -10
$$
You replaced the $?$ with $0$, automatically assuming that. If you automatically assume that, of course you can show it (kinda redundant). If you continued to use $?$ you should get:
$$
-10b = -10+?
$$
The $?$ is an unknown that's not given to you, just like $a$ and $b$, so you treat it like another variable, such as $c$. Now that you have $3$ variables in a system of $2$ linear equations you can easily see why there is no solution.
In summary, $? = c$, $?\ne 0$.
A: Looking at it from a geometric pespective, we have (a,b) lies in the line $3x+5y=2$. According to the question we have to find the the value of $15a+15b$. Hence we have to complete the equation of the line $15x+15y=c$. This implies that we have to find the point of intersection of the two lines which is not possible.
A: You've find a solution imposing $?=0$ but you could decide that $?=10$ and find $a=\frac{2}{3}, b=0$, the system cannot be solved because the value of $a$ and $b$ depends on that of $?$
