Linear Systems 4D6F2B

1. **Problem Statement:** Solve a system of linear equations. 2. **General Formula:** A linear system can be written as: $$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$ where $a_1, b_1, c_1, a_2, b_2, c_2$ are constants. 3. **Methods to Solve:** - Substitution - Elimination - Matrix methods (e.g., inverse matrix, Cramer's rule) 4. **Example:** Solve $$\begin{cases} 2x + 3y = 6 \\ 4x - y = 5 \end{cases}$$ 5. **Step 1: Use elimination to eliminate $y$:** Multiply the second equation by 3: $$4x - y = 5 \implies 12x - 3y = 15$$ 6. **Step 2: Add the first equation and the new equation:** $$2x + 3y = 6$$ $$12x - 3y = 15$$ Adding: $$2x + 3y + 12x - 3y = 6 + 15$$ $$14x + \cancel{3y} - \cancel{3y} = 21$$ 7. **Step 3: Simplify and solve for $x$:** $$14x = 21$$ $$x = \frac{21}{14} = \frac{3}{2}$$ 8. **Step 4: Substitute $x = \frac{3}{2}$ into the first equation:** $$2\left(\frac{3}{2}\right) + 3y = 6$$ $$3 + 3y = 6$$ $$3y = 6 - 3 = 3$$ $$y = \frac{3}{3} = 1$$ 9. **Final answer:** $$x = \frac{3}{2}, \quad y = 1$$ This means the solution to the system is the point $\left(\frac{3}{2}, 1\right)$ where both lines intersect.