Does anyone know if number #7 is a solution or no solution and does anyone know how to do #10 and #6

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  7. (x, y) = (0, 1)

  10. (x, y) ={((7+√97)/2, (-1-√97)/2) and ((7-√97)/2, (-1+√97)/2)

  6. (x, y) = (-3+√6, -1+√6) and (-3-√6, -1-√6)

Step-by-step explanation:

7. It is convenient to subtract the second equation from the first:

  (y) -(y) = (x^2 +5x +1) -(x^2 +2x +1)

  0 = 3x . . . . simplify

  0 = x . . . . . divide by 3

  y = 0 + 0 + 1 . . . . substitute x=0 into either equation

The solution is (x, y) = (0, 1). The first attachment shows this solution.

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10. It can work to subtract the first equation from the second.

  (y) -(x +y) = (x^2 -8x -9) -(3)

  -x = x^2 -8x -12 . . . . simplify

  x^2 -7x = 12 . . . . . . . add x+12, swap sides

  x^2 -7x +49/4 = 12 +49/4 . . . . complete the square

  (x -7/2)^2 = 97/4 . . . . . . . . write as a square

  x -7/2 = (±√97)/2 . . . . . . .  square root

  x = (7±√97)/2 . . . . . add 7/2

  y = 3 -x = (-1±√97)/2 . . . . find corresponding y

Solutions are (x, y) = {((7+√97)/2, (-1-√97)/2), ((7-√97)/2, (-1+√97)/2)}. The second attachment shows the solution.

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6. For this one, it can work well to subtract the first equation from the second.

  (y) -(y) = (x +2) - (-x^2 -5x -1)

  0 = x^2 +6x +3

Adding 6 to both sides completes the square.

  x^2 +6x +9 = 6

  (x +3)^2 = 6

  x +3 = ±√6

  x = -3±√6

  y = x +2 = -1±√6

Solutions are (x, y) = (-3+√6, -1+√6) and (-3-√6, -1-√6). The third attachment shows the solution.

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About completing the square

In general, completing the square starts with ...

  x^2 +ax = b

Then the square is completed by adding the square of a/2.

  x^2 +ax +(a/2)^2 = b + (a/2)^2

Then you can write the left side as a square.

  (x +a/2)^2 = (b +a^2/4)

Finally, take the square root and subtract the left-side constant.

  x = -a/2 ± √(b +a^2/4)

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step-by-step explanation: