1. **Problem 1 (Easy):** Solve for $x$ in the equation $$2x + 3 = 7$$. 2. **Problem 2 (Medium):** Find the roots of the quadratic equation $$x^2 - 5x + 6 = 0$$. 3. **Problem 3 (Hard):** Solve the system of equations: $$\begin{cases} 2x + y = 5 \\ 3x - 2y = 4 \end{cases}$$. --- ### Step-by-step solutions: **1. Solve $2x + 3 = 7$** 1. Start with the equation: $$2x + 3 = 7$$. 2. Subtract 3 from both sides: $$2x = 7 - 3$$. 3. Simplify the right side: $$2x = 4$$. 4. Divide both sides by 2: $$x = \frac{4}{2}$$. 5. Simplify: $$x = 2$$. **Answer:** $x = 2$ --- **2. Find roots of $x^2 - 5x + 6 = 0$** 1. The quadratic equation is $$x^2 - 5x + 6 = 0$$. 2. Use the factoring method: find two numbers that multiply to 6 and add to -5. 3. These numbers are -2 and -3. 4. Factor the quadratic: $$(x - 2)(x - 3) = 0$$. 5. Set each factor equal to zero: - $$x - 2 = 0 \Rightarrow x = 2$$ - $$x - 3 = 0 \Rightarrow x = 3$$ **Answer:** $x = 2$ or $x = 3$ --- **3. Solve the system:** $$\begin{cases} 2x + y = 5 \\ 3x - 2y = 4 \end{cases}$$ 1. From the first equation, express $y$ in terms of $x$: $$y = 5 - 2x$$. 2. Substitute $y$ into the second equation: $$3x - 2(5 - 2x) = 4$$. 3. Distribute: $$3x - 10 + 4x = 4$$. 4. Combine like terms: $$7x - 10 = 4$$. 5. Add 10 to both sides: $$7x = 14$$. 6. Divide both sides by 7: $$x = 2$$. 7. Substitute $x = 2$ back into $y = 5 - 2x$: $$y = 5 - 2(2) = 5 - 4 = 1$$. **Answer:** $x = 2$, $y = 1$