Найти наименьшее и наибольшее значение функции z=z(x, y) в замкнутой области D. z=x^2+xy-2 D: y=4x^2-4 , y<=0

5 11/14 х - 8/15=5/21                                    
5 11/14 х=5/21+ 8/15
5 11/14 х=  (25+56)/105
 81/14 х= 81/105
х = 81/105 :81/14
х= (81*14)/(105*81)
х= 14/105=2/15

7 3/10+ 25/28 х =8 13/35
25/28 х  = 293/35-73/10
25/28 х  = 293/35-73/10
25/28 х  = (586-511)/70
25/28 х  = 75/70
х = (75*28)/(70*25)
х = 6/5 = 1 1/5 = 1,2

3 1\3 - 1 1/20 х = 1 14/15
-21/20 х = 29/15-10/3
-21/20 х = (29-50)/15
21/20 х = 21/15
х = (21*20)/(15*21) = 4/3 = 1 1/3

3/8 х + 7/12 х - 5/6 х = 9/32
((3*3+7*2-5*4)/24) х = 9/32
3/24 х = 9/32
1/8 х = 9/32
х = 9/32 : 1/8
х = (9*8)/(32*1) = 9/4 = 2 1/4 = 2,25

2 1\3 : х - 1 1/6 = 1 5/9
7/3 :х = 14/9 + 7/6
7/3 : х = (28+21)/18
7/3 : х = 49/18
х = 7/3 : 49/18 = (7*18)/(3*49) = 7/7

2 1/3 :( х - 1 1/6) = 1 5/9
7/3 : х - 7/3 : 7/6 = 14/9
7/3 : х - 2 = 14/9
7/3 : х = (14+18)/9
х = 7/3 : 32/9 = (7*9)/(3*32) = 21/32