I've got a simultaneous equation, and have attempted it. Hope you guys could check my answer out.
$\begin{cases}9x-y=-1~......(1)\\-6x+3y=10~......(2)\end{cases}$
$(1)\times3$ :
$27x-3y=3~......(3)$
$(3)-(2)$ :
$21x=-7$
$3x=-1$
2026-05-17 07:44:14.1779003854
Simultaneous equation difficulty minus
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From the second two equations: Yes, $21x = -7 \iff 3x = -1\iff x = -\frac 13$. Now solve for $y$.
In the first two equations, you should have $27 x = 7 \iff 3x = 1 \iff x = \frac 13$. Now solve for $y$.
ALERT: Note that the two systems you posted are identical, save for the equation $27 - 3y = 3$ in the second "pair". I think you must have multiplying the first equation by $3$, but dropped a "sign". You should have gotten $$27 - 3y = -3,$$ in which case you'll find that $x = \frac 13$ is the solution for $x$..