Chapter 8 Systems of Equations and Inequalities

2
2
2
2
x + xy = 6
⎛ y + 3⎞ ⎜ ⎟
⎝
2y ⎠
⎛ y + 3⎞
+ ⎜ ⎟y=
6
⎝
2y
⎠
4 2 2
y + 6y + 9 y + 3
+ = 6
2
4y
2
y + 6y + 9+ 2y + 6y = 24y
4 2
3y − 12y + 9= 0
4 2
y − 4y + 3= 0
2
y −3 2
y − 1 = 0
4 2 4 2 2
( )( )
Thus, y =± 3 or y =± 1 .
If y = 1: x = 2 ⋅ 1 = 2
If y =− 1: x = 2( − 1) =− 2
If y = 3 : x = 3
If y =− 3 : x =− 3
Solutions: (2, 1), (–2, −1), ( 3, 3 ) , ( − 3, − 3)
⎧ 2 2
x −xy− y =
⎪ 2 0
48. ⎨
⎪⎩ xy + x + 6= 0
Factor the first equation, solve for x, substitute
into the second equation and solve:
2 2
x −xy− 2y = 0
( x− 2 y)( x+ y)
= 0
x = 2 y or x =−y
Substitute x = 2y
and solve:
xy + x + 6= 0
(2 yy ) + 2y =−6
2
2y + 2y+ 6= 0
2
2( y + y+
3) = 0
y =
2
− 1± 1 −4(1)(3)
(No real solution)
2(1)
Substitute x =− y and solve:
xy + x + 6= 0
−y⋅ y+ ( − y)
= −6
2
−y − y+
6= 0
( −y−3)( y−
2) = 0
y =−3
or y=2
If y =− 3 : x = 3
If y = 2 : x = −2
Solutions: (3, –3), (– 2, 2)
867
Section 8.6: Systems of Nonlinear Equations
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49.
⎧ 2 2
+ + − − =
y y x x 2 0
⎪
⎨
x − 2
⎪ y + 1+ = 0
⎩
y
Multiply each side of the second equation by –y
and add the equations to eliminate y:
2 2
y + y+ x −x− 2= 0
2
−y − y − x+
2 = 0
2
x − 2x = 0
x x−
2 = 0
( )
x = 0 or x = 2
If x = 0 :
2 2
y + y+ 0 −0− 2= 0 ⇒
2
y + y−
2= 0
⇒ ( y+ 2)( y− 1) = 0⇒ y =− 2 or y = 1
If x = 2 :
2 2
y + y+ 2 −2− 2= 0 ⇒
2
y + y = 0
⇒ yy ( + 1) = 0⇒ y= 0 or y=−1
Note: y ≠ 0 because of division by zero.
Solutions: (0, –2), (0, 1), (2, –1)
⎧ 3 2 2
x − x + y + y−
=
⎪
2
50. 2 3 4 0
⎨ y − y
⎪ x − 2+ = 0
2
⎩
x
Multiply each side of the second equation by
2
− x and add the equations to eliminate x:
3 2 2
x − 2x + y + 3y− 4= 0
3 2 2
− x + 2 x − y + y = 0
4y − 4 = 0
4y = 4
y = 1
If y = 1:
3 2 2 3 2
x − 2x + 1 + 3⋅1− 4= 0 ⇒ x − 2x = 0
2
⇒ x ( x− 2) = 0⇒ x = 0 or x = 2
Note: x ≠ 0 because of division by zero.
Solution: (2, 1)