Question

fie ne a,b,c daca
a=9*7-2*4-5*10
b=98-(7*7+3*7+5*3)
c=(13+7*4)-39
calculeaza
2a+3c-b
b+4a-2c
(a+c)*9-b

Answer 1

a = 9 x7 - 2 x4 - 5 x10
a = 63 - 8 - 50
a = 5

b = 98 - ( 7 x7 + 3x7 + 5x3 )
b = 98 - (49 + 21 + 15)
b = 98 - 85
b = 13

c =( 13 + 7 x4 ) - 39
c = 13 + 28 - 39
c = 2

2a + 3c - b = 2 x5 + 3 x 2 - 13 = 10 + 6 - 13 = 16 - 13 = 3
b + 4a - 2c = 13 + 4 x 5 - 2 x2 = 13 + 20 - 4 = 33 - 4 = 29
( a + c ) x 9 - b = ( 5 + 2 ) x 9 - 13 = 7 x 9 - 13 = 63 - 13 =50

Answer 2

$a=9*7-2*4-5*10 \\ a=63-8-50 \\ a=55-50 \\ a=5 \\ b=98-(7*7+3*7+5*3) \\ b=98-(49+21+15) \\ b=98-(70+15) \\ b=98-85 \\ b=13 \\ c=(13+7*4)-39 \\ c=(13+28)-39 \\ c=41-39 \\ c=2$
$a=5;b=13;c=2 \\ 2a+3c-b=2*5+3*2-13=10+6-13=16-13=3 \\ b+4a-2c=13+4*5-2*2=13+20-4=33-4=29 \\ (a+c)*9-b=(5+2)*9-13=7*9-13=63-13=50$