Simultaneous Equations Df076D

1. **State the problem:** We need to solve the simultaneous equations: $$2x + y = 5$$ $$3x - \sqrt{2}y = 7$$ and express $x$ and $y$ in the form $a + b\sqrt{2}$ where $a,b$ are integers. 2. **Write down the system:** $$\begin{cases} 2x + y = 5 \\ 3x - \sqrt{2}y = 7 \end{cases}$$ 3. **Express $y$ from the first equation:** $$y = 5 - 2x$$ 4. **Substitute $y$ into the second equation:** $$3x - \sqrt{2}(5 - 2x) = 7$$ 5. **Expand and simplify:** $$3x - 5\sqrt{2} + 2x\sqrt{2} = 7$$ 6. **Group terms with $x$:** $$3x + 2x\sqrt{2} = 7 + 5\sqrt{2}$$ 7. **Factor out $x$:** $$x(3 + 2\sqrt{2}) = 7 + 5\sqrt{2}$$ 8. **Solve for $x$ by dividing:** $$x = \frac{7 + 5\sqrt{2}}{3 + 2\sqrt{2}}$$ 9. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate $3 - 2\sqrt{2}$: $$x = \frac{(7 + 5\sqrt{2})(3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}$$ 10. **Calculate denominator:** $$(3)^2 - (2\sqrt{2})^2 = 9 - 8 = 1$$ 11. **Calculate numerator:** $$7 \times 3 = 21$$ $$7 \times (-2\sqrt{2}) = -14\sqrt{2}$$ $$5\sqrt{2} \times 3 = 15\sqrt{2}$$ $$5\sqrt{2} \times (-2\sqrt{2}) = -10 \times 2 = -20$$ Sum numerator terms: $$21 - 14\sqrt{2} + 15\sqrt{2} - 20 = (21 - 20) + (-14\sqrt{2} + 15\sqrt{2}) = 1 + \sqrt{2}$$ 12. **So:** $$x = 1 + \sqrt{2}$$ 13. **Find $y$ using $y = 5 - 2x$:** $$y = 5 - 2(1 + \sqrt{2}) = 5 - 2 - 2\sqrt{2} = 3 - 2\sqrt{2}$$ 14. **Final answers:** $$x = 1 + \sqrt{2}$$ $$y = 3 - 2\sqrt{2}$$ These are in the form $a + b\sqrt{2}$ with integers $a,b$ as required.