A Smarter way to solve this system of linear equations? I am a high school student and when practicing for the SAT I stumbled across this question:

$$
\begin{eqnarray}
−0.2x + by &=& 7.2\\
5.6x − 0.8y​ &=& 4
\end{eqnarray}
$$
Consider the system of equations above. For what value of $b$ will
the system have exactly one solution $(x,y)$ with $x=2$? Round the
answer to the nearest tenth.

My initial though was that if the x don't cancel each other by elimination then I must find out a way to make y cancel out. So I directly put $b = 0.8$ without really thinking about it.
But when I had a look at the answer sheet they solved it by finding the value of y from the second equation then replacing the value y they got in the first equation to get $b$. By doing so they were able to get $b = 0.8$ just like I did.
So my question is do I really need to follow their time consuming way, or will my way of solving work of all questions of this type?
 A: Unfortunately, I'd say that your method of simply guessing b = 0.8 to make things convenient for you was simply luck. That is most certainly not always the case.
My advice: carefully study their solution and learn how they did it. Reattempt the question using their method and use their method for all future questions. You got lucky guessing the value of b = 0.8, it is not always clear cut, and often the questions are not made to be convenient, so your method will not work.
You could, of course, trial and error values of b. But for a majority of questions this would be far more time consuming and frustrating.
A: Since it says "exactly one solution $(x,y)$ with $x=2$" you can first insert $x=2$":
$$
\begin{eqnarray}
−0.2 \times 2 + by &=& 7.2\\
5.6 \times 2 − 0.8y​ &=& 4
\end{eqnarray}
$$
The second equation gives $y=(5.6 \times 2 - 4)/0.8 = 9.$ Then the first equation gives $b = (7.2 + 0.2 \times 2)/9 = 0.8444\ldots \approx 0.8$.
A: They are telling you that
$$\begin{align}by&=7.2+0.2\cdot2=7.6,\\-0.8y&=4-5.6\cdot2=-7.2\end{align}.$$
Then
$$b=0.8\frac{7.6}{7.2}=0.8\frac{19}{18}.$$
