Question 179666
Three times a number plus twice a second number is 41. Also, four times the first plus five times the second is 71. Find the number
:
let a and b be our numbers
:
3a+2b=41.........eq 1
4a+5b=71.........eq 2
:
multiply eq 1 by -4 and eq 2 by 3 and add the two equations together
:
-4(3a+2b=41)---->-12a-8b=-164
3(4a+5b=71)------>12a+15b=213
:
-12a-8b=-164.....eq 1 revised
12a+15b=213......eq 2 revised
:
the a terms are eliminated -12a+12a=0. We are left with -8b+15b=-164+213
:
7b=49
:
{{{highlight(b=7)}}}
:
take b's value and plug it back into any equation. I chose eq 1
:
3a+2(7)=41
:
3a=27
:
{{{highlight(a=9)}}}