To eliminate the $y$-variable, we'll multiply the first equation by $-\frac{11}{3}$.

Given the first equation: $2x + y = 3$

Now, multiply both sides by $-\frac{11}{3}$:

$-\frac{11}{3} \times (2x + y) = -\frac{11}{3} \times 3$

Distribute $-\frac{11}{3}$ to both terms on the left side: $-\frac{11}{3} \times 2x - \frac{11}{3} \times y = -\frac{11}{3} \times 3$ This simplifies to: $-\frac{22}{3}x - \frac{11}{3}y = -11$

So, the new equation is: $-\frac{22}{3}x - \frac{11}{3}y = -11$

Now, we have the modified version of the first equation. Do you want me to proceed with solving the system from here? Let me know if you'd like any more details or clarifications.

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Related Questions:

1. How do we combine the two equations to eliminate $y$?
2. What happens if we multiply by a different factor to eliminate $y$?
3. How do we solve for $x$ after eliminating $y$?
4. Can this method be applied to systems of equations with different coefficients?
5. How do we check the solution of a system of equations using substitution?

Tip: Always check your work when performing multiplication on both sides of an equation—especially with fractions!