Step 1: Solve the system algebraically first, by elimination.
We are given the two equations:
\[ x - 2y = 3 \quad \text{(Equation 1)} \]
\[ 3x - 8y = 7 \quad \text{(Equation 2)} \]
Instead of only plotting points and reading off the crossing point from the graph, we first pin down exactly where the lines should cross, using elimination, so we know precisely which point to expect on the graph.
Multiply Equation 1 by 3, so the coefficient of $x$ matches Equation 2:
\[ 3(x - 2y) = 3(3) \implies 3x - 6y = 9 \quad \text{(Equation 3)} \]
Step 2: Eliminate x by subtracting.
Subtract Equation 2 from Equation 3:
\[ (3x - 6y) - (3x - 8y) = 9 - 7 \]
\[ -6y + 8y = 2 \]
\[ 2y = 2 \implies y = 1 \]
Substitute $y = 1$ back into Equation 1:
\[ x - 2(1) = 3 \implies x = 5 \]
So the algebraic solution is $x = 5$, $y = 1$, meaning the lines must cross exactly at the point $(5, 1)$.
Step 3: Build the point tables for plotting, using this known crossing point as a guide.
For Equation 1, $x - 2y = 3$, rewrite as $x = 2y + 3$ and pick convenient values of $y$:
If $y = 0$: $x = 3$, giving point $(3, 0)$.
If $y = 1$: $x = 5$, giving point $(5, 1)$, the expected crossing point.
If $y = -1$: $x = 1$, giving point $(1, -1)$.
For Equation 2, $3x - 8y = 7$, rewrite as $x = \frac{8y + 7}{3}$ and pick values of $y$ that keep $x$ a whole number:
If $y = 1$: $x = \frac{15}{3} = 5$, giving point $(5, 1)$, matching the same expected point.
If $y = -2$: $x = \frac{-9}{3} = -3$, giving point $(-3, -2)$.
Step 4: Draw the lines and confirm the crossing.
Plot the points $(3, 0)$, $(5, 1)$, $(1, -1)$ and join them with a straight line for Equation 1. Plot the points $(5, 1)$ and $(-3, -2)$ and join them with a straight line for Equation 2.
Both lines pass through $(5, 1)$, and since two distinct straight lines cross at only one point, this confirms the crossing point matches the algebraic answer found in Step 2.
Final Answer:
\[ \boxed{x = 5, \ y = 1} \]