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To solve this system of linear equations, we have:

\[
5x - 3y = 18
\]
\[
2x - 3y = 3
\]

To eliminate \( y \), we can subtract the second equation from the first:

\[
(5x - 3y) - (2x - 3y) = 18 - 3
\]

Simplifying this gives:

\[
5x - 3y - 2x + 3y = 15
\]

The \( y \) terms cancel out, leaving:

\[
3x = 15
\]

Now, solve for \( x \) by dividing both sides by 3:

\[
x = \frac{15}{3} = 5
\]

Now that we know \( x = 5 \), we can substitute this value back into one of the original equations to solve for \( y \). Let’s use the second equation:

\[
2x - 3y = 3
\]

Substitute \( x = 5 \):

\[
2(5) - 3y = 3
\]
\[
10 - 3y = 3
\]

Now, isolate \( y \):

\[
-3y = 3 - 10
\]
\[
-3y = -7
\]
\[
y = \frac{-7}{-3} = \frac{7}{3}
\]

Thus, the solution to the system in fraction form is:

\[
(x, y) = \left( 5, \frac{7}{3} \right)
\]

Would you like further details or have any questions?

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Here are five additional questions to explore:

1. How would the solution change if both equations had different coefficients for \( y \)?
2. What happens if we solve the system by substitution instead of elimination?
3. Can we solve this system graphically and what would that look like?
4. How would the solution differ if the equations were inconsistent (no solution)?
5. What if the system was dependent (infinitely many solutions)?

**Tip:** When solving systems, try elimination if the coefficients of one variable match or can easily be made to match, as it often simplifies calculations.

Solve the system for (x, y). Fraction form only.
5x - 3y = 18
2x - 3y = 3

Learn how to solve a system of linear equations using the elimination method. This problem involves solving for x and y in fraction form in the system 5x - 3y = 18 and 2x - 3y = 3. Suitable for students in Grades 8-10.

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