1. The problem is to solve the system of equations by eliminating $y$. 2. Suppose the system is: $$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$ 3. To eliminate $y$, multiply each equation by a suitable number so that the coefficients of $y$ are opposites. 4. For example, multiply the first equation by $b_2$ and the second by $-b_1$: $$\begin{cases} b_2(a_1x + b_1y) = b_2c_1 \\ -b_1(a_2x + b_2y) = -b_1c_2 \end{cases}$$ which simplifies to $$\begin{cases} a_1b_2x + b_1b_2y = b_2c_1 \\ -a_2b_1x - b_1b_2y = -b_1c_2 \end{cases}$$ 5. Add the two equations to eliminate $y$: $$ (a_1b_2 - a_2b_1)x + (b_1b_2 - b_1b_2)y = b_2c_1 - b_1c_2 $$ which reduces to $$ (a_1b_2 - a_2b_1)x = b_2c_1 - b_1c_2 $$ 6. Solve for $x$: $$ x = \frac{b_2c_1 - b_1c_2}{a_1b_2 - a_2b_1} $$ 7. Substitute $x$ back into one of the original equations to find $y$. This method eliminates $y$ and finds $x$ first, then $y$.