Asked by Grant
Eureka is 16 years younger than Renton. In 3 years Renton will be twice as old as Eureka. How old are they now?
Eureka's Age:
Renton's Age:
Eureka's Age:
Renton's Age:
Answers
Answered by
Bosnian
E = Eureka's Age
R = Renton's Age
E = R - 16
In 3 years Renton will be R + 3
Eureca will be E + 3
In 3 years Renton will be twice as old as Eureka.
This mean :
R + 3 = 2 ( E + 3 )
R + 3 = 2 ( R - 16 + 3 )
R + 3 = 2 ( R - 13 )
R + 3 = 2 R - 26 Add 26 to both sies
R + 3 + 26 = 2 R - 26 + 26
R + 29 = 2 R Subtract R to both sides
R + 29 - R = 2 R - R
29 = R
R = 29
E = R - 16
E = 29 - 16
E = 13
Eureka's Age = 13
Renton's Age = 29
Proof :
In 3 years Renton will be 32
Eureca will be 16
32 / 16 = 2
Renton will be twice as old as Eureka.
R = Renton's Age
E = R - 16
In 3 years Renton will be R + 3
Eureca will be E + 3
In 3 years Renton will be twice as old as Eureka.
This mean :
R + 3 = 2 ( E + 3 )
R + 3 = 2 ( R - 16 + 3 )
R + 3 = 2 ( R - 13 )
R + 3 = 2 R - 26 Add 26 to both sies
R + 3 + 26 = 2 R - 26 + 26
R + 29 = 2 R Subtract R to both sides
R + 29 - R = 2 R - R
29 = R
R = 29
E = R - 16
E = 29 - 16
E = 13
Eureka's Age = 13
Renton's Age = 29
Proof :
In 3 years Renton will be 32
Eureca will be 16
32 / 16 = 2
Renton will be twice as old as Eureka.
Answered by
Graham
Let y be Eureka's age in years.
Let x be Renton's age in years.
So:
y = x-16
2(y+3) = x+3
Rearrange to express x as a function of y.
x = y + 16
x = 2y + 3
Eliminating x by equating:
y + 16 = 2y +3
=> y = 13
Substituting back to obtain x.
x = 13 + 16
=> x = 29
Let x be Renton's age in years.
So:
y = x-16
2(y+3) = x+3
Rearrange to express x as a function of y.
x = y + 16
x = 2y + 3
Eliminating x by equating:
y + 16 = 2y +3
=> y = 13
Substituting back to obtain x.
x = 13 + 16
=> x = 29
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