1. 2x-y=2 3x+y=8 2. 4x+y=3 3x-y-11 3. 4x-3y=3 4x-7y=-5

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вопрос

To solve these systems of equations, we can use either substitution or elimination method. Let's solve each system separately:

Let's add the two equations together to eliminate y:

2x - y + 3x + y = 2 + 8
5x = 10
x = 2

Now, substitute x = 2 back into one of the equations to solve for y:
2(2) - y = 2
4 - y = 2
y = 2

Therefore, the solution to the first system is x = 2, y = 2.

Let's add the two equations together to eliminate y:
4x + y + 3x - y = 3 + 11
7x = 14
x = 2

Now, substitute x = 2 back into one of the equations to solve for y:
4(2) + y = 3
8 + y = 3
y = -5

Therefore, the solution to the second system is x = 2, y = -5.

Let's subtract the second equation from the first to eliminate x:
(4x - 3y) - (4x - 7y) = 3 - (-5)
4y = 8
y = 2

Now, substitute y = 2 back into one of the equations to solve for x:
4x - 3(2) = 3
4x - 6 = 3
4x = 9
x = 9/4

Therefore, the solution to the third system is x = 9/4, y = 2.