Answer:
The first score is 109
Step-by-step explanation:
I am assuming that in the first sentence of the question, you meant:
Sum of the students three scores is 231...
First, let the scores of the first second and third student be a, b and c respectively. We are told that:
a + b + c = 231 . . . . . . . .(1) (sum of students three scores is 231)
a = b + 20 . . . . . . . . . . . (2) (the first is 20 points more than the second)
a + b = 6c . . . . . . . . . . . .(3) (sum of the first two is 6 more times the third)
required, find a.
substituting the value of (a + b) in equation (3) into equation (1), we will have the following:
since a + b = 6c . . . (3)
a + b + c = 231 . . . . . (1), becomes,
(a + b) + c = 231
(6c) + c = 231
7c = 231 (divide both sides by 7)
c = 231 ÷ 7 = 33
∴ c = 33
Next, from equation (2), we know that a = b + 20; this can also be written as:
a - 20 = b
∴ b = a - 20 . . . . . . . (4)
Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:
a + (a - 20) + 33 = 231
(a + a) - 20 + 33 = 231
2a + 13 = 231
2a = 231 - 13 = 218
a = 218 ÷ 2 = 109
∴ a = 109
we can also calculate for 'b' by substituting for the value of 'a' in equation 4
b = a - 20 = 109 - 20 = 89.
and to test if the values of a, b and c are correct:
a + b + c = 231
109 + 89 + 33 = 231
![60x^3+145x^2+70x=5x(12x^2+29x+14)=5x(12x^2+21x+8x+14)\\\\=5x[3x(4x+7)+2(4x+7)]=5x(4x+7)(3x+2)](https://tex.z-dn.net/?f=60x%5E3%2B145x%5E2%2B70x%3D5x%2812x%5E2%2B29x%2B14%29%3D5x%2812x%5E2%2B21x%2B8x%2B14%29%5C%5C%5C%5C%3D5x%5B3x%284x%2B7%29%2B2%284x%2B7%29%5D%3D5x%284x%2B7%29%283x%2B2%29)
