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Answer:
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)
