Finding X or Y first in simultaneous equation

Question

I have the question "Solve the pair of simultaneous equations"

$$\begin{cases}3X + 3Y + 4 = 0\\ 5X - 2Y - 5 = 0\end{cases}$$

When I checked the solutions it states that we need to make the $Y$ values the same so that they cancel out and we do this by multiplying the first equation by $2$ and the second equation by $3$.

Could you rather multiply the first equation by $5$ and the second by $3$ to make the both $X$'s the same and cancel the $X$'s instead ? When I did this however I got two different answers.

Answer 1

This also works and yields the correct solution.

$$\begin{align*} 15X + 15Y + 20 &= 0 \\ 15X - 6Y - 15 &= 0 \\[.5cm] 21Y + 35 &= 0 \\ 15X - 6Y - 15 &= 0 \\[.5cm] Y &= -\frac{35}{21} = -\frac{5}{3} \\ 15X &= 6Y + 15 = -6\cdot\frac{5}{3} + 15 = 5\\[.5cm] Y &= -\frac{5}{3} \\ X &= \frac{5}{15} = \frac{1}{3} \end{align*}$$

Answer 2

One thing to take care is that in the two equations the coefficients of X and Y are of opposite signs. If we want to cancel the Y's then we have to multiply the first equation by 2 and the second by 3 and then ADD them. If we want to cancel the X's then we have to multiply the first equation by 5 and the second one by 3 and then SUBTRACT them. This is one precaution to take care of.

Answer 3

Note that the answer depends on the field $K$. For coding theory and other applications we might want to solve these two equation over finite fields, for example, or for a field of prime characteristic $p>0$. Then, for characteristic $p=3$ we have no solution, because the first equation would be $0=1$, a contradiction. For $p=5$ we obtain $2Y=0$, hence $Y=0$ since $2\neq 0$ for $p=5$. Then $X=2$. So in all other cases, we obtain $X=3^{-1}$ and $Y=-5\cdot 3^{-1}$ as above (which is also true for $p=5$).

So your question "Could you rather multiply the first equation by 5 and the second by 3" then has not always a positive answer.