the sum of two numbers is 80 and the greater number exceeds twice the smaller by 11 find the number
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Answer:
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Step-by-step explanation:
Solution:-
Solution:-Let the two numbers be x and y such that x>y..
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=80
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=80
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y=
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 3
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23Substituting the value of y in eq
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23Substituting the value of y in eq n
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23Substituting the value of y in eq n (2), we have
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23Substituting the value of y in eq n (2), we havex=2×23+11
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23Substituting the value of y in eq n (2), we havex=2×23+11⇒x=46+11=57
Solution:-Let the two numbers be x and y such that x>y..Now according to the question-Condition I:-x+y=80.....(1)Condition II:-x=2y+11.....(2)Substituting the value of x in eq n (1) we have(2y+11)+y=803y+11=803y=80−11y= 369 =23Substituting the value of y in eq n (2), we havex=2×23+11⇒x=46+11=57Hence the required numbers are 57 and 23.
Answer:
let the numbers be x and 2x +11
x +2x+11=80
3x= 80-11
3x= 69
x=69/3
x= 23
Step-by-step explanation:
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