Solve the system of equations.%0D%0A2x+8y=8%0D%0A2%0D%0A%0D%0A+%0D%0A8%0D%0A%0D%0A=%0D%0A8%0D%0A%0D%0A2x−4y=−16%0D%0A2%0D%0A%0D%0A−%0D%0A4%0D%0A%0D%0A=%0D%0A−%0D%0A16%0D%0A(1 point)%0D%0A(, )

Answer 1

The system of equations is:

2x + 8y = 8 .................(1)
2x - 4y = -16 .................(2)

We can solve this system of equations by either substitution or elimination method.

Using elimination method:
Multiply equation (2) by 2 to make the coefficients of x in both equations the same:
2(2x - 4y) = 2(-16)
4x - 8y = -32 .................(3)

Now, subtract equation (3) from equation (1) to eliminate y:
(2x + 8y) - (4x - 8y) = 8 - (-32)
2x + 8y - 4x + 8y = 8 + 32
-2x + 16y = 40 .................(4)

Next, divide equation (4) by 2 to simplify:
(-2x + 16y)/2 = 40/2
- x + 8y = 20 .................(5)

Now we have two equations:
- x + 8y = 20 .................(5)
2x + 8y = 8 .................(1)

Add equation (5) to equation (1) to eliminate x:
(-x + 8y) + (2x + 8y) = 20 + 8
x + 16y = 28 .................(6)

Now we have two equations:
x + 16y = 28 .................(6)
2x + 8y = 8 .................(1)

We can subtract equation (6) from equation (1) to eliminate x:
(2x + 8y) - (x + 16y) = 8 - 28
x - 8y = -20 .................(7)

Now we have two equations:
x - 8y = -20 .................(7)
x + 16y = 28 .................(6)

Adding equations (7) and (6), we can eliminate x:
(x - 8y) + (x + 16y) = -20 + 28
2x + 8y = 8 .................(1)

This is the same equation as equation (1), which means the system of equations is dependent and has infinitely many solutions. Therefore, there are no unique values of x and y to be determined.