1. The problem is to solve a system of linear equations using the elimination method. 2. The elimination method involves adding or subtracting the equations to eliminate one variable, making it easier to solve for the other. 3. Suppose the system is: $$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$ 4. Multiply one or both equations by suitable numbers so that the coefficients of one variable are opposites. 5. Add the equations to eliminate that variable. 6. Solve the resulting single-variable equation. 7. Substitute the found value back into one of the original equations to find the other variable. 8. Example: $$\begin{cases} 2x + 3y = 8 \\ 4x - 3y = 2 \end{cases}$$ Multiply the first equation by 1 and the second by 1 (already set): $$\begin{cases} 2x + 3y = 8 \\ 4x - 3y = 2 \end{cases}$$ Add the two equations: $$ (2x + 3y) + (4x - 3y) = 8 + 2 $$ $$ 2x + 4x + 3y - 3y = 10 $$ $$ 6x + \cancel{3y} - \cancel{3y} = 10 $$ $$ 6x = 10 $$ Divide both sides by 6: $$ \frac{6x}{\cancel{6}} = \frac{10}{6} $$ $$ x = \frac{5}{3} $$ Substitute $x=\frac{5}{3}$ into the first equation: $$ 2\left(\frac{5}{3}\right) + 3y = 8 $$ $$ \frac{10}{3} + 3y = 8 $$ Subtract $\frac{10}{3}$ from both sides: $$ 3y = 8 - \frac{10}{3} $$ $$ 3y = \frac{24}{3} - \frac{10}{3} = \frac{14}{3} $$ Divide both sides by 3: $$ y = \frac{14}{3} \times \frac{1}{3} = \frac{14}{9} $$ Final solution: $$ x = \frac{5}{3}, \quad y = \frac{14}{9} $$