1. **Problem:** Find the solution set of the system of equations: $$\begin{cases} x - y = 1 \\ 5x^2 + 2y^2 = 53 \end{cases}$$ 2. **Step 1: Express $y$ in terms of $x$ from the first equation.** From $x - y = 1$, we get: $$y = x - 1$$ 3. **Step 2: Substitute $y = x - 1$ into the second equation.** $$5x^2 + 2(x - 1)^2 = 53$$ 4. **Step 3: Expand and simplify.** $$(x - 1)^2 = x^2 - 2x + 1$$ So, $$5x^2 + 2(x^2 - 2x + 1) = 53$$ $$5x^2 + 2x^2 - 4x + 2 = 53$$ $$7x^2 - 4x + 2 = 53$$ 5. **Step 4: Bring all terms to one side.** $$7x^2 - 4x + 2 - 53 = 0$$ $$7x^2 - 4x - 51 = 0$$ 6. **Step 5: Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=7$, $b=-4$, $c=-51$. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 7 \times (-51) = 16 + 1428 = 1444$$ $$\sqrt{1444} = 38$$ So, $$x = \frac{4 \pm 38}{14}$$ 7. **Step 6: Find the two values of $x$.** - For $+$: $$x = \frac{4 + 38}{14} = \frac{42}{14} = 3$$ - For $-$: $$x = \frac{4 - 38}{14} = \frac{-34}{14} = -\frac{17}{7}$$ 8. **Step 7: Find corresponding $y$ values using $y = x - 1$.** - For $x=3$: $$y = 3 - 1 = 2$$ - For $x = -\frac{17}{7}$: $$y = -\frac{17}{7} - 1 = -\frac{17}{7} - \frac{7}{7} = -\frac{24}{7}$$ 9. **Final solution set:** $$\boxed{\{(3, 2), \left(-\frac{17}{7}, -\frac{24}{7}\right)\}}$$ This means the system has two solutions: $(3, 2)$ and $\left(-\frac{17}{7}, -\frac{24}{7}\right)$.