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r + 2t = -3 . . . . (A)

3r - 4t = -9 . . . . (B)

2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9

5r = - 15 so r = -3

substituting in (A), t = 0

So the answer is (r, t) = (-3, 0)

r + 2t = -3 . . . . (A)

3r - 4t = -9 . . . . (B)

2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9

5r = - 15 so r = -3

substituting in (A), t = 0

So the answer is (r, t) = (-3, 0)

r + 2t = -3 . . . . (A)

3r - 4t = -9 . . . . (B)

2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9

5r = - 15 so r = -3

substituting in (A), t = 0

So the answer is (r, t) = (-3, 0)

r + 2t = -3 . . . . (A)

3r - 4t = -9 . . . . (B)

2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 9

5r = - 15 so r = -3

substituting in (A), t = 0

So the answer is (r, t) = (-3, 0)

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Q: Which pair of values is the solution to the system of equations below r plus 2t equals -3 and 3r-4t equals -9?

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