Problem: Solve $7x – 5y = 2$ and $x + 2y = 3$.

Method 1: The Substitution Method

This method works best when one of the variables is “easy” to isolate (meaning it has a coefficient of 1).

1. Label your equations:

   • (1) $7x – 5y = 2$

   • (2) $x + 2y = 3$

2. Isolate one variable: From equation (2), we can easily solve for $x$:
   $$x = 3 – 2y$$
3. Substitute into the other equation: Plug this new expression for $x$ into equation (1):
   $$7(3 – 2y) – 5y = 2$$

4. Solve for $y$: * Distribute: $21 – 14y – 5y = 2$

   • Combine like terms: $21 – 19y = 2$

   • Isolate $y$: $-19y = -19 \implies \mathbf{y = 1}$

5. Find $x$: Plug $y = 1$ back into our isolated $x$ equation:
   $$x = 3 – 2(1) \implies \mathbf{x = 1}$$

Final Answer: $(x, y) = (1, 1)$

Method 2: The Elimination Method

The goal here is to manipulate the equations so that when you add or subtract them, one variable completely disappears (is “eliminated”).

1. Align the equations:

   • (1) $7x – 5y = 2$

   • (2) $x + 2y = 3$

2. Create a match: To eliminate $x$, multiply equation (2) by 7 so it matches the $7x$ in equation (1).

   • Equation (2) becomes: $7x + 14y = 21$

3. Subtract to eliminate: Subtract the new equation (2) from equation (1):
   $$(7x – 5y) – (7x + 14y) = 2 – 21$$
   $$-19y = -19 \implies \mathbf{y = 1}$$
4. Solve for $x$: Substitute $y = 1$ into original equation (2):
   $$x + 2(1) = 3 \implies \mathbf{x = 1}$$

Final Answer: $(x, y) = (1, 1)$

Method 3: The Cross-Multiplication Method

The Cross-Multiplication Method is a systematic way to solve linear equations by using their coefficients directly. It is particularly useful for avoiding long algebraic manipulations.

Problem: Solve $7x – 5y – 2 = 0$ and $x + 2y – 3 = 0$.

1. Arrange the Coefficients: Write down the coefficients of $y$, the constants, and the coefficients of $x$ in this specific sequence:

   • Row 1: $-5, -2, 7, -5$

   • Row 2: $2, -3, 1, 2$

2. Apply the Cross-Multiplication Formula: Follow the downward and upward diagonal multiplications: 
   $$\frac{x}{(-5)(-3) – (2)(-2)} = \frac{y}{(-2)(1) – (-3)(7)} = \frac{1}{(7)(2) – (1)(-5)}$$

3. Simplify the Denominators:

   • $\frac{x}{15 + 4} = \frac{y}{-2 + 21} = \frac{1}{14 + 5}$

   • $\frac{x}{19} = \frac{y}{19} = \frac{1}{19}$

4. Find the Final Values: Solve each ratio against the constant:

   • $x = \frac{19}{19} = 1$

   • $y = \frac{19}{19} = 1$

Solution: $(x, y) = (1, 1)$.

Practice: Handling Fractions with Substitution

Real-world math isn’t always clean integers. Let’s look at an example where the solution results in fractions.

Problem: Solve $3x + 2y = 5$ (1) and $2x – 3y = 7$ (2).

1. Isolate a Variable: From equation (2), isolate $x$:

   • $2x = 7 + 3y \implies x = \frac{7 + 3y}{2}$

2. Substitute and Solve: Plug this expression for $x$ into equation (1):

   • $3(\frac{7+3y}{2}) + 2y = 5$

   • Multiply by 2 to clear the fraction: $3(7 + 3y) + 4y = 10$

   • $21 + 9y + 4y = 10 \implies 13y = -11$

   • $y = -\frac{11}{13}$

3. Find the second value: Substitute $y$ back into the simplified $x$ equation:

   • $3x = 5 + \frac{22}{13}$

   • $3x = \frac{65 + 22}{13} \implies 3x = \frac{87}{13}$

   • $x = \frac{29}{13}$

Final Answer: $(x, y) = (\frac{29}{13}, -\frac{11}{13})$.