Determine whether the following systems of equations have one solution, no solution, or infinitely many solutions: 7. x + y = 50 and 2x + 5y = 160, x = 30 and y = 20. 8. y = (1/4)x... Determine whether the following systems of equations have one solution, no solution, or infinitely many solutions: 7. x + y = 50 and 2x + 5y = 160, x = 30 and y = 20. 8. y = (1/4)x - 2 and 8y - 2x = -16. 9. 2x + 4y = -6 and (1/2)x + y = 3, 12 = -6. Read more

Steps to Solve

1. Check the solution for system 7

Substitute $x = 30$ and $y = 20$ into the equations $x + y = 50$ and $2x + 5y = 160$.

For $x + y = 50$: $30 + 20 = 50$, which is true.

For $2x + 5y = 160$: $2(30) + 5(20) = 60 + 100 = 160$, which is also true.

Since the given values of $x$ and $y$ satisfy both equations, this system has one solution.

2. Analyze system 8

The second equation $8y - 2x = -16$ can be simplified by dividing by 2 to get $4y - x = -8$. Solving for $y$, we have $4y = x - 8$, and hence $y = \frac{1}{4}x - 2$. This is the same as the first equation. Since the two equations are equivalent, and the result is $0 = 0$, this system has infinitely many solutions.

3. Analyze system 9

The result of operations on the given equations is $12 = -6$, which is false. This indicates that the two equations are inconsistent and there is no solution to the system.