i set the equation end give you the solutions but you try to get there
x=first test, y=second test z= three test
i)second test score was 8 points higher than her first score => y=x+8
ii)Her third score was 88=> z=88
iii)B average (between 80=> B=(x+y+z)/3<80
iiii)B average...and 89 inclusive)=> B=(x+y+z)/3>89
advice: replace i and ii in disequations iii and iiii
solution x>72 x<85,50
Briana's second test score was 8 points higher than her first score. Her third score was 88. She had a B average (between 80 and 89 inclusive) for three tests. What can you conclude about her first test score?
2 answers
sorry.....errata corrige:
iii)B average (between 80=> B=(x+y+z)/3>80
iiii)B average...and 89 inclusive)=> B=(x+y+z)/3<89
iii)B average (between 80=> B=(x+y+z)/3>80
iiii)B average...and 89 inclusive)=> B=(x+y+z)/3<89