SOLUTION: I need help with this problem: Three times Nancy's age is 5 more than twice Rogers age. The sum of their age is 32. find their ages.

Question 269473: I need help with this problem: Three times Nancy's age is 5 more than twice Rogers age. The sum of their age is 32. find their ages.
Found 2 solutions by oberobic, JBarnum:
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
n = Nancy's age
r = Roger's age
.
We are told:
n = 2r + 5
n + r = 32
.
Viewing as a simultaneous equations:
n -2r = 5
n + r = 32
Subtracting the second equation from the first:
-3r = -27
Dividing by -3
r = 9
Roger's age is 9.
.
substituting back into the sum:
n + r = 32
n + 9 = 32
n = 32 - 9
n = 23
Nancy's age is 23.
.
So the sum of their ages is 32.
.
But is Nancy's age 5 more than twice Roger's?
.
2(9) + 5 = 23
That checks, so all we have to do is state our answer:
Nancy is 23 years old.
Roger is 9 years old.
.
Done.
Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
"Three times Nancy's age is 5 more than twice Rogers age." means 3N=5+2R
"The sum of their age is 32." means N+R=32
N=nancys age and R=rogers age
3N-2R=5
1N+1R=32 times this whole line by 2
_________________________
3N-2R=5
2N+2R=64 add the 2 lines together vertically
________
5N=69
N=69/5=13.8
Nancy is 13.8yrs old
or 13years and 292days old
___________
so if N+R=32
13.8+R=32
R=18.2
Roger is 18.2yrs old
or 18years and 73days old
___________________
check
3N=5+2R
3(13.8)=5+2(18.2)
41.4=41.4
correct
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