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let x = score on second exam and y = score on third exam
so x would be 67
we know that "The second score was 7 less than the third score", so we can make this equation: x = y - 7
67 is the first exam score
the second may be 67, but we haven't gotten there yet
so would it be like this 67x=y-7
so we know this so far first exam score: 67 second exam score: x third exam score: y replace x with y-7 (since x = y-7) to get first exam score: 67 second exam score: y-7 third exam score: y
now you add up all 3 scores, then divide by 3 to get the average of 84 like this (Score1 + Score2 + Score3)/3 = Average (67 + (y-7) + y)/3 = 84 (67 + y-7 + y)/3 = 84 (2y+60)/3 = 84 2y+60 = 84*3 ... ... ... y = ???
oh ok so i use this equation right ?:)
kk one sec
so you don't have to divide the 3?
what do you mean
(2y+60)/3 = 84 2y+60 = 84*3 like here how instead of dividing you multiplied 3 to 84
oh i multiplied both sides by 3 because in the end you want to isolate y
so you undo that division by 3
oh ok :) thnks one moment
is it 96 ?
yep, y = 96 now find x
okay now do you plug in the y to the very first equation you gave me?
yep x = y - 7
is it 89
yep x = 89 y = 96
yesss! thank you so mcuh :)
so going back to what we defined x and y to be (remember at the top we said "let x = score on second exam and y = score on third exam", so you can see how it all comes around full circle) we know first exam score = 67 (given) second exam score = 89 (just found) third exam score = 96 (just found)