SOLUTION: the sum ot three numbers is 105. the third is 11 less than ten times the second. twice the first is 7 more than three times the second. find the numbers.
my work so far:
a+b+
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Question 98445This question is from textbook elementary and intermediate algebra
: the sum ot three numbers is 105. the third is 11 less than ten times the second. twice the first is 7 more than three times the second. find the numbers.
my work so far:
a+b+c=105
10b=c+11 >> 10b-c=1
2a=7+3b >> 2a-3b=7
next i know i have to get rid of one of the numbers so i decided on "a" and i started with equations 1 and 3 so i multiplyed 1 by 2 so the "a"s would cancel out.
-2a-2b-c=105
+
2a-3b=7
-5b-c=112
but now im stuck i dont know what to do next.
This question is from textbook elementary and intermediate algebra
Answer by mathslover(157) (Show Source): You can put this solution on YOUR website!
Let the second number be x
from the problem statement "the third is 11 less than ten times the second"
third number = 10x -11
again from the problem statement "first is 7 more than three times the second"
first number = 3x + 7
Sum of the numbers =105
3x+7 + x + 10x-11 =105
14x = 109
x= 109/14
second number is 109/14
first number = 10* 109/14 -11
=936/14
third number = 3* 109/14 + 7
=425/14
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