Solve the following system of linear equations using (a) substitution and (b) determinants. \[ \begin{array}{r}3 x+5 y-13 \\ x-2 y=0.75\end{array} \]

Let's solve the system of linear equations using both the **substitution method** and the **determinant (Cramer's) method**. The given system is: \[ \begin{cases} 3x + 5y = 13 \\ x - 2y = 0.75 \end{cases} \] --- ### **(a) Substitution Method** 1. **Solve the second equation for \( x \):** \[ x = 0.75 + 2y \] 2. **Substitute \( x = 0.75 + 2y \) into the first equation:** \[ 3(0.75 + 2y) + 5y = 13 \] 3. **Expand and simplify:** \[ 2.25 + 6y + 5y = 13 \\ 11y = 13 - 2.25 \\ 11y = 10.75 \\ y = \frac{10.75}{11} = \frac{43}{44} \] 4. **Find \( x \) using the expression from step 1:** \[ x = 0.75 + 2\left(\frac{43}{44}\right) \\ x = \frac{3}{4} + \frac{86}{44} = \frac{33}{44} + \frac{86}{44} = \frac{119}{44} \] **Solution:** \[ x = \frac{119}{44}, \quad y = \frac{43}{44} \] --- ### **(b) Determinant (Cramer's) Method** 1. **Write the system in matrix form:** \[ \begin{cases} 3x + 5y = 13 \\ x - 2y = 0.75 \end{cases} \] 2. **Compute the main determinant \( D \):** \[ D = \begin{vmatrix} 3 & 5 \\ 1 & -2 \end{vmatrix} = 3(-2) - 5(1) = -6 - 5 = -11 \] 3. **Compute \( D_x \) by replacing the first column with the constants:** \[ D_x = \begin{vmatrix} 13 & 5 \\ 0.75 & -2 \end{vmatrix} = 13(-2) - 5(0.75) = -26 - 3.75 = -29.75 = -\frac{119}{4} \] 4. **Compute \( D_y \) by replacing the second column with the constants:** \[ D_y = \begin{vmatrix} 3 & 13 \\ 1 & 0.75 \end{vmatrix} = 3(0.75) - 13(1) = 2.25 - 13 = -10.75 = -\frac{43}{4} \] 5. **Find \( x \) and \( y \) using Cramer's Rule:** \[ x = \frac{D_x}{D} = \frac{-\frac{119}{4}}{-11} = \frac{119}{44} \] \[ y = \frac{D_y}{D} = \frac{-\frac{43}{4}}{-11} = \frac{43}{44} \] **Solution:** \[ x = \frac{119}{44}, \quad y = \frac{43}{44} \] --- Both methods yield the same solution: \[ \boxed{\left( \dfrac{119}{44},\; \dfrac{43}{44} \right)} \]