These are 3 seperate questions i need with.

Step-by-step explanation:

1.  Subtract the first equation from twice the second to eliminate x2.

  2(2x1 -x2) -(x1 -2x2) = 2(-2) -(-7)

  3x1 = 3 . . . . . simplify

  x1 = 1 . . . . . . .divide by 3

Substitute this value into the second equation.

  2·1 -x2 = -2

  4 -x2 = 0 . . . . . add 2

  4 = x2 . . . . . . . add x2

The solution is (x1, x2) = (1, 4).

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2. Multiply the first equation by 3 and add the second equation.

  3(-5x1 -3x2) +(15x1 +9x2) = 3(7) +(2)

  0 = 23 . . . . . not true;

There is no solution.

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3. There are two equations in 3 unknowns, so there cannot be a unique solution. The equations are not dependent, so there will be an infinite number of solutions that can be written in terms of a single parameter (t).

Let x3 = t. Then the system of equations can be rewritten as

  2x1 - 5x2 = 23 +3t

  x1 - 4x2 = 16

Subtracting twice the second equation from the first, we have ...

  (2x1 -5x2) -2(x1 -4x2) = (23 +3t) -2(16)

  3x2 = 3t -9 . . . simplify

  x2 = t -3 . . . . . divide by 3

Substituting this into the second equation above, we have ...

  x1 -4(t -3) = 16

  x1 = 16 +4t -12 . . . . . . add 4(t-3)

  x1 = 4t +4 . . . . . . . . simplify

The solutions are (x1, x2, x3) = (4t+4, t-3, t).

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The attachment shows the first system solved by graphical means. This validates our answer.