Question

Roliver is three times as old as Cerise. Three years ago, the sum of their ages was sixty-six. Find their present age?

Answer 1

Cerise is currently 18 and Roliver is 54.

Explanation:

Let Cerise's current age be `x`,
If Roliver is currently three times her age, his current age is thus `3x`.

So 3 years ago, Cerise must have been `x-3` years old and Roliver must have been `3x-3` years old.

Since we know that the sum of their ages 3 years ago was 66, we can form an equation:

`(3x-3) + (x-3) = 66`

`4x=72`

`x=18`

Hence, Cerise's age is `18` and Roliver's is `3(18) = 54`.

Answer 2

Cerise's current age is 12
Roliver's current age is 36

Explanation:

Let Roliver's current age be `r`
Let Cerise's current age be `c`

Breaking the question down into its component parts

Roliver is `............................ r=?`
tree timeas as old as Cerise `... r=3c`
Three years ago `...................r-3=3c-3" ".. Equation(1)`

The sum of their ages was`..(r-3)+(3c-3)=?`
sixty six `...................(r-3)+(3c-3)=66" "..Equation(2)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider `Eqn(1)`

Add 3 to both sides

`color(green)(r=3c...Equation(1_a))`

Consider `Eqn(2)` and write as:

`color(green)(r+3c-6=66)`

Add 6 to both sides

`color(green)(r+3c=72...Equation(2_a)`

Using `Eqn(1_a)` substitute for `r` in `Eqn(2_a)`

`color(green)(color(red)(r)+3c=72color(white)("ddd") ->color(white)("ddd") color(red)(3c)+3c=72`

`color(green)(color(white)("dddddddddddd")->color(white)("ddd")6c=72`

Divide both sides by 6

`color(green)(color(white)("ddddddd")->color(white)("d")bar(ul(|color(white)(./.) c=72/6=12color(white)(./.)|))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute for `c` in `Eqn(1_a)`

`color(green)(r=3color(red)(c)color(white)("ddd") ->color(white)("ddd") r=3xxcolor(red)(72/6) =36 )`

`color(white)("ddddddddddddd")color(green)(bar(ul(|color(white)(./.)r=36color(white)(./.)|)))`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

check

`Eqn(1)`

`r-3=3c-3color(white)("dddd") ->color(white)("dddd") 36-3=3(12)-3`

`Eqn(2)`

`(r-3)+(3c-3)=66color(white)("ddd")->color(white)("ddd") 36-3+3(12)-3 = 66`