Elena is now four times as old as Lizette. Five years ago, the sum of their ages was 45. How old is each now?

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Answer 1 · 2017-06-14T12:40:53

Elenea is $44$ $44$ years old and Lizette $11$ $11$ .

Call the present ages of Elena and Lizette $e and l$ $e and l$ ; you get:

$e = 4 l$ $e=4l$

$(e - 5) + (l - 5) = 45$ $(e-5)+(l-5)=45$

Substitute the first into the second and rearrange:

$4 l - 5 + l - 5 = 45$ $4l-5+l-5=45$

$5 l = 55$ $5l=55$

$l = 11$ $l=11$

so that

$e = 4 \cdot 11 = 44$ $e=4*11=44$

Answer 2 · 2017-06-14T12:59:14

Lizette is $11$ $11$ and Elena is $4 \times 11 = 44$ $4xx11 = 44$

There are 2 people and 2 periods of time (past and present)

Choose a variable and write an expression for the age of each person for each period of time..

Lizette is younger, so let her age be $x$ $x$ years.
Elena is four times as old, so her age is $4 x$ $4x$ years

$5$ $5$ years ago they were both $5$ $5$ years younger.

$color(white)(xxxxxxx)ul("(5 years ago"color(white)(xxxxx)"present")$

Lizette $color(white)(xxxxx)color(blue)((x-5))color(white)(xxxxxxxx)x$
Elena $color(white)(xxxxx)color(blue)((4x-5))color(white)(xxxvvvxx)4x$

The sum of their ages $5$ years ago was $45$ years.

Write an equation to show this:

$color(blue)((x-5)+(4x-5) = 45)$

$5x -10 = 45$

$5x = 45+10$

$5x = 55$

$x=11$

So, Lizette is $11$ and Elena is $4xx11 = 44$

Check: $5$ years ago: Lizette was $6$ and Elena was $39$

$6+39 = 45$

It all checks out!.