1. **Problem Statement:** Solve the system of linear equations by substitution and addition (elimination) methods. Since the system of equations is not provided, let's consider a general example: $$\begin{cases} 2x + 3y = 6 \\ x - y = 1 \end{cases}$$ --- 2. **Substitution Method:** - Step 1: Solve one equation for one variable. From the second equation: $$x = y + 1$$ - Step 2: Substitute this expression for $x$ into the first equation: $$2(y + 1) + 3y = 6$$ - Step 3: Simplify and solve for $y$: $$2y + 2 + 3y = 6$$ $$5y + 2 = 6$$ $$5y = 4$$ $$y = \frac{4}{5}$$ - Step 4: Substitute $y$ back into $x = y + 1$: $$x = \frac{4}{5} + 1 = \frac{9}{5}$$ --- 3. **Addition (Elimination) Method:** - Step 1: Write the system: $$\begin{cases} 2x + 3y = 6 \\ x - y = 1 \end{cases}$$ - Step 2: Multiply the second equation by 3 to align $y$ coefficients: $$3(x - y) = 3(1) \Rightarrow 3x - 3y = 3$$ - Step 3: Add this to the first equation: $$ (2x + 3y) + (3x - 3y) = 6 + 3 $$ $$ 5x = 9 $$ $$ x = \frac{9}{5} $$ - Step 4: Substitute $x$ into the second original equation: $$ \frac{9}{5} - y = 1 $$ $$ -y = 1 - \frac{9}{5} = -\frac{4}{5} $$ $$ y = \frac{4}{5} $$ --- **Final Answer:** $$x = \frac{9}{5}, \quad y = \frac{4}{5}$$