Solve the system of equations 2x+8y=8 and 2x-4y=-16

Answer 1

To solve the system of equations 2x + 8y = 8 and 2x - 4y = -16, we can proceed using the method of elimination.

First, let's multiply the second equation by 2 to make the coefficients of x in both equations the same:

2(2x - 4y) = 2(-16)
4x - 8y = -32

Now, we can subtract the first equation from the second equation to eliminate x:

(4x - 8y) - (2x + 8y) = -32 - 8
4x - 8y - 2x - 8y = -40
2x - 16y = -40

Now we have a new equation: 2x - 16y = -40

Let's multiply the first equation by 2 to make the coefficients of x the same in both equations:

2(2x + 8y) = 2(8)
4x + 16y = 16

Now, let's add the equations:

(4x + 16y) + (2x - 16y) = 16 - 40
4x + 16y + 2x - 16y = -24
6x = -24

Divide both sides of the equation by 6:

6x/6 = -24/6
x = -4

Now we can substitute this value of x into one of the original equations to solve for y. Let's use the first equation:

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

Divide both sides of the equation by 8:

8y/8 = 16/8
y = 2

So the solution to the system of equations 2x + 8y = 8 and 2x - 4y = -16 is x = -4 and y = 2.