To answer this question, we can use the substitution method or the elimination method.
If we use the substitution method, we can proceed as follows:
1. Solve the first equation for x:

If we add 5y to both sides of the equation, we have:

2. We can substitute the corresponding value of x - as a function of y - as follows:

3. Now, we can solve the equation for y as follows:
a. Apply the distributive property:

b. Adding like terms, and subtracting 4 from both sides:

4. If we substitute this value for y in the first equation, we will have:

Therefore, we have that the solution for this system of linear equations is x = 2, and y = 0.
We can check these results, if we substitute those values into the original equations:
x = 2
y = 0


In summary, we have that the solution to the system of equations is:
x = 2 and y = 0.