The system of linear equations shown in the image is:

$3x - 14y = 0$ $x - 4y + 1 = 0$

Let's solve this system of equations step-by-step.

Step 1: Solve the second equation for $x$.

$x - 4y + 1 = 0$ Rearrange it to express $x$ in terms of $y$: $x = 4y - 1$

Step 2: Substitute $x = 4y - 1$ into the first equation.

Substitute this expression for $x$ into the first equation $3x - 14y = 0$: $3(4y - 1) - 14y = 0$

Step 3: Simplify and solve for $y$.

Expand the equation: $12y - 3 - 14y = 0$ Combine like terms: $-2y - 3 = 0$ Solve for $y$: $-2y = 3$ $y = -\frac{3}{2}$

Step 4: Substitute $y = -\frac{3}{2}$ into the expression for $x$.

Substitute $y = -\frac{3}{2}$ into $x = 4y - 1$: $x = 4\left(-\frac{3}{2}\right) - 1$ $x = -6 - 1 = -7$

Solution:

The solution to the system of equations is: $x = -7, \quad y = -\frac{3}{2}$

Let me know if you would like more details or if you have any questions!

Here are 5 related questions you might want to explore:

1. How can we verify the solution of a system of equations?
2. Can all systems of linear equations be solved using substitution?
3. What happens if the system has no solution or infinitely many solutions?
4. How can this system be solved using the elimination method?
5. How does the graphical method work for solving a system of linear equations?

Tip: When solving systems of equations, always substitute your solution back into the original equations to verify its correctness.