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?

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

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

To determine Briana's first test score, let's work through the given information step by step.

First, we know that her second test score was 8 points higher than her first score. Let's call her first score F, and her second score S. So, we can express this relationship as:

S = F + 8

Second, we are told that her third test score was 88. Let's call her third score T. So, we have:

T = 88

Third, we are informed that Briana had a B average for the three tests. A B average falls between 80 and 89 inclusive. Therefore, to find the average of the three tests, we sum them up and divide by 3:

(Average of three tests) = (F + S + T) / 3

Now, we can substitute the given values into this equation:

(F + (F + 8) + 88) / 3 = (F + (F + 8) + 88) / 3

Simplifying the equation further:

(F + F + 8 + 88) / 3 = (2F + 96) / 3

Since Briana's average falls within the B range, we know that:

80 ≤ (2F + 96) / 3 ≤ 89

To solve this inequality, we can multiply all parts of the inequality by 3 to eliminate the denominator:

240 ≤ 2F + 96 ≤ 267

Next, we subtract 96 from all parts of the inequality:

144 ≤ 2F ≤ 171

Finally, dividing all parts of the inequality by 2:

72 ≤ F ≤ 85.5

Therefore, based on the B average condition, we can conclude that Briana's first test score falls between 72 and 85.5.