I'm terrible at pre calc, could someone tell me if the solution set for log subscript 2 x + log sub 2 (x-1)=1 is the solution -1? Using either the elimination or substitution method is the solution set for 4x + y =11 and 2x -y = 7 is the answer (3, -1)? Kudos!

Yes.

To find the solution set for the equation log₂x + log₂(x - 1) = 1, you can use the properties of logarithms.

1. Combine the logarithms: log₂[x(x - 1)] = 1
2. Convert the equation into exponential form: 2^1 = x(x - 1)
3. Simplify the left side: 2 = x(x - 1)
4. Expand and rearrange the equation: x² - x - 2 = 0
5. Factoring the quadratic equation: (x - 2)(x + 1) = 0
6. Set each factor equal to zero: x - 2 = 0 or x + 1 = 0
7. Solve for x: x = 2 or x = -1

So, the solution set for the equation is {2, -1}. Therefore, -1 is indeed one of the solutions.

Coming to the second equation set: 4x + y = 11 and 2x - y = 7.

1. You can solve this system of equations using the elimination method.
2. Multiply the second equation by 2: 4x + y = 11 and 4x - 2y = 14.
3. Subtract the second equation from the first equation: (4x + y) - (4x - 2y) = 11 - 14.
4. Simplify and combine like terms: 4x + y - 4x + 2y = -3.
5. Simplify further: 3y = -3.
6. Solve for y: y = -1.
7. Plug the value of y into the first equation: 4x + (-1) = 11.
8. Simplify and solve for x: 4x - 1 = 11, 4x = 12, x = 3.

Therefore, the solution set for the system of equations is (x, y) = (3, -1). So, your answer of (3, -1) is indeed correct. Kudos to you!