1. **State the problem:** Solve the simultaneous equations: $$x^2 + xy - 4y^2 = 2$$ $$2x - 3y = 1$$ 2. **Use the second equation to express $x$ in terms of $y$: ** From $$2x - 3y = 1$$, add $3y$ to both sides: $$2x = 1 + 3y$$ Divide both sides by 2: $$x = \frac{1 + 3y}{2}$$ Note: You cannot cancel the 2 with the 1 or 3y separately; the entire numerator is divided by 2. 3. **Substitute $x = \frac{1 + 3y}{2}$ into the first equation:** $$\left(\frac{1 + 3y}{2}\right)^2 + \left(\frac{1 + 3y}{2}\right) y - 4y^2 = 2$$ 4. **Expand and simplify:** Square the first term: $$\frac{(1 + 3y)^2}{4} + \frac{(1 + 3y) y}{2} - 4y^2 = 2$$ Expand $(1 + 3y)^2 = 1 + 6y + 9y^2$: $$\frac{1 + 6y + 9y^2}{4} + \frac{y + 3y^2}{2} - 4y^2 = 2$$ 5. **Multiply through by 4 to clear denominators:** $$1 + 6y + 9y^2 + 2(y + 3y^2) - 16y^2 = 8$$ 6. **Expand and combine like terms:** $$1 + 6y + 9y^2 + 2y + 6y^2 - 16y^2 = 8$$ Combine $y$ terms: $$6y + 2y = 8y$$ Combine $y^2$ terms: $$9y^2 + 6y^2 - 16y^2 = (15 - 16) y^2 = -y^2$$ So the equation becomes: $$1 + 8y - y^2 = 8$$ 7. **Bring all terms to one side:** $$-y^2 + 8y + 1 - 8 = 0$$ Simplify constants: $$-y^2 + 8y - 7 = 0$$ Multiply both sides by $-1$ to make the leading coefficient positive: $$\cancel{-}y^2 + 8y - 7 = 0 \Rightarrow y^2 - 8y + 7 = 0$$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times 7}}{2 \times 1} = \frac{8 \pm \sqrt{64 - 28}}{2} = \frac{8 \pm \sqrt{36}}{2}$$ $$y = \frac{8 \pm 6}{2}$$ So two solutions for $y$: $$y = \frac{8 + 6}{2} = 7$$ $$y = \frac{8 - 6}{2} = 1$$ 9. **Find corresponding $x$ values using $x = \frac{1 + 3y}{2}$:** For $y=7$: $$x = \frac{1 + 3 \times 7}{2} = \frac{1 + 21}{2} = \frac{22}{2} = 11$$ For $y=1$: $$x = \frac{1 + 3 \times 1}{2} = \frac{1 + 3}{2} = \frac{4}{2} = 2$$ 10. **Final solutions:** $$(x, y) = (11, 7) \quad \text{or} \quad (2, 1)$$