1. **Problem Statement:** Solve the system of simultaneous equations. 2. **Two-variable linear system:** Solve $$\begin{cases} 2x + 3y = 6 \\ x - y = 1 \end{cases}$$ 3. **Formula and rules:** Use substitution or elimination method to solve linear systems. 4. **Step-by-step solution for two-variable system:** From second equation: $$x = y + 1$$ Substitute into first: $$2(y + 1) + 3y = 6$$ Simplify: $$2y + 2 + 3y = 6$$ $$5y + 2 = 6$$ $$5y = 6 - 2$$ $$5y = 4$$ $$y = \frac{4}{5}$$ Substitute back: $$x = \frac{4}{5} + 1 = \frac{9}{5}$$ 5. **Three-variable linear system:** Solve $$\begin{cases} x + y + z = 6 \\ 2x - y + 3z = 14 \\ -x + 4y - z = -2 \end{cases}$$ 6. **Method:** Use elimination or substitution to reduce variables stepwise. 7. **Step-by-step:** Add first and third: $$ (x - x) + (y + 4y) + (z - z) = 6 + (-2) $$ $$ 0 + 5y + 0 = 4 $$ $$ 5y = 4 $$ $$ y = \frac{4}{5} $$ Substitute $y$ into first: $$ x + \frac{4}{5} + z = 6 $$ $$ x + z = 6 - \frac{4}{5} = \frac{30}{5} - \frac{4}{5} = \frac{26}{5} $$ Substitute $y$ into second: $$ 2x - \frac{4}{5} + 3z = 14 $$ $$ 2x + 3z = 14 + \frac{4}{5} = \frac{70}{5} + \frac{4}{5} = \frac{74}{5} $$ Now solve system: $$ \begin{cases} x + z = \frac{26}{5} \\ 2x + 3z = \frac{74}{5} \end{cases} $$ Multiply first by 2: $$ 2x + 2z = \frac{52}{5} $$ Subtract from second: $$ (2x + 3z) - (2x + 2z) = \frac{74}{5} - \frac{52}{5} $$ $$ z = \frac{22}{5} $$ Substitute back: $$ x + \frac{22}{5} = \frac{26}{5} $$ $$ x = \frac{26}{5} - \frac{22}{5} = \frac{4}{5} $$ 8. **Non-linear and linear system:** Solve $$\begin{cases} y = x^2 + 1 \\ y = 3x + 1 \end{cases}$$ 9. **Method:** Set equations equal since both equal $y$: $$ x^2 + 1 = 3x + 1 $$ Simplify: $$ x^2 = 3x $$ $$ x^2 - 3x = 0 $$ Factor: $$ x(x - 3) = 0 $$ Solutions: $$ x = 0 \quad \text{or} \quad x = 3 $$ Find $y$ for each: For $x=0$: $$ y = 3(0) + 1 = 1 $$ For $x=3$: $$ y = 3(3) + 1 = 10 $$ **Final answers:** - Two-variable system: $x=\frac{9}{5}$, $y=\frac{4}{5}$ - Three-variable system: $x=\frac{4}{5}$, $y=\frac{4}{5}$, $z=\frac{22}{5}$ - Non-linear and linear system: $(0,1)$ and $(3,10)$