1. **Stating the problem:** We are given the system of linear equations: $$3x + 4y = 15$$ $$5x - 2y = -1$$ We need to find the values of $x$ and $y$ using two methods: substitution or elimination, and determinants. 2. **Method a: Substitution or Elimination** We will use the elimination method here. 3. Multiply the first equation by 2 and the second equation by 4 to align coefficients of $y$: $$2(3x + 4y) = 2(15) \Rightarrow 6x + 8y = 30$$ $$4(5x - 2y) = 4(-1) \Rightarrow 20x - 8y = -4$$ 4. Add the two equations to eliminate $y$: $$6x + 8y + 20x - 8y = 30 - 4$$ $$26x = 26$$ 5. Solve for $x$: $$x = \frac{26}{26} = 1$$ 6. Substitute $x=1$ into the first original equation to find $y$: $$3(1) + 4y = 15 \Rightarrow 3 + 4y = 15$$ $$4y = 15 - 3 = 12$$ $$y = \frac{12}{4} = 3$$ 7. **Method b: Determinants (Cramer's Rule)** The system can be written as: $$\begin{cases} 3x + 4y = 15 \\ 5x - 2y = -1 \end{cases}$$ 8. Calculate the determinant $D$ of the coefficient matrix: $$D = \begin{vmatrix} 3 & 4 \\ 5 & -2 \end{vmatrix} = (3)(-2) - (4)(5) = -6 - 20 = -26$$ 9. Calculate determinant $D_x$ by replacing the first column with constants: $$D_x = \begin{vmatrix} 15 & 4 \\ -1 & -2 \end{vmatrix} = (15)(-2) - (4)(-1) = -30 + 4 = -26$$ 10. Calculate determinant $D_y$ by replacing the second column with constants: $$D_y = \begin{vmatrix} 3 & 15 \\ 5 & -1 \end{vmatrix} = (3)(-1) - (15)(5) = -3 - 75 = -78$$ 11. Solve for $x$ and $y$: $$x = \frac{D_x}{D} = \frac{-26}{-26} = 1$$ $$y = \frac{D_y}{D} = \frac{-78}{-26} = 3$$ 12. **Final answer:** $$x = 1, \quad y = 3$$ This is the point where the two lines intersect.