Solve the linear equations using the methods provided in the problem set.

Steps to Solve

1. Solve for y in equation 1

We have $4x + 5y = 19$, $x = 1$. Substitute the value of $x$ into the equation and solve for $y$: $4(1) + 5y = 19$ $4 + 5y = 19$ $5y = 19 - 4$ $5y = 15$ $y = \frac{15}{5} = 3$

2. Solve for x in equation 2

We have $2x + 5y = 16$, $y = 2$. Substitute the value of $y$ into the equation and solve for $x$: $2x + 5(2) = 16$ $2x + 10 = 16$ $2x = 16 - 10$ $2x = 6$ $x = \frac{6}{2} = 3$

3. Calculate $D_x$ using Cramer's rule

For the system $3x + 2y = 11$ and $7x - 4y = 9$, $D_x$ is given by the determinant of the matrix formed by replacing the x coefficients with the constants: $D_x = \begin{vmatrix} 11 & 2 \ 9 & -4 \end{vmatrix} = (11 \times -4) - (2 \times 9) = -44 - 18 = -62$

4. Calculate $D_y$ using Cramer's rule

For the system $3x + y = 1$ and $2x - 11y = 3$, $D_y$ is given by the determinant of the matrix formed by replacing the y coefficients with the constants: $D_y = \begin{vmatrix} 3 & 1 \ 2 & 3 \end{vmatrix} = (3 \times 3) - (1 \times 2) = 9 - 2 = 7$

5. Solve for x in equation 5

We have $3x - 2y = -1$, $y = 2$. Substitute the value of $y$ into the equation and solve for $x$: $3x - 2(2) = -1$ $3x - 4 = -1$ $3x = -1 + 4$ $3x = 3$ $x = \frac{3}{3} = 1$

6. Find the intersection point of the two lines

We have the system $x + 3y = 7$ and $2x + y = -1$. We can multiply the second equation by -3 to eliminate $y$: $-3(2x + y) = -3(-1) \implies -6x - 3y = 3$ Now add this modified equation to the first equation: $(x + 3y) + (-6x - 3y) = 7 + 3$ $-5x = 10$ $x = \frac{10}{-5} = -2$ Substitute $x = -2$ into the first equation to solve for $y$: $(-2) + 3y = 7$ $3y = 7 + 2$ $3y = 9$ $y = \frac{9}{3} = 3$ So, the point of intersection is $(-2, 3)$.

7. Find the value of k for parallel lines

For the lines $14x + 10y = 33$ and $kx + 5y = 11$ to be parallel, the ratio of the coefficients of $x$ and $y$ must be equal, but not equal to the ratio of the constants. Thus: $\frac{14}{k} = \frac{10}{5}$ $10k = 14 \times 5$ $10k = 70$ $k = \frac{70}{10} = 7$ We need to check that $\frac{14}{7} = \frac{10}{5} \neq \frac{33}{11}$ $2 = 2 \neq 3$ The condition is satisfied.