Question

Rene is $6$ years older than her younger sister.

After $10$ years, the sum of their ages will be $50$ years.

How do you find their present ages?

Answer 1

Find their present ages:

Consider the present age of Rene's younger sister be $x$ and the Rene present age be$x+6$.
After ten years, the sum of their ages will be equal to $50$ years.

According to the given information:

$\begin{array}{rcl}\left[\left(x+6\right)+10\right]+\left(x+10\right)& =& 50\\ & \Rightarrow & \left(x+16\right)+\left(x+10\right)=50\\ & \Rightarrow & 2x+26=50\\ & \Rightarrow & 2x=50-26\\ & \Rightarrow & 2x=24\\ & \Rightarrow & x=12\end{array}$

The age of Rene's younger sister $\left(x\right)$ is $12$ years and Rene's age $\left(x+6\right)$ is $18$ years

Hence, the present ages of Rene and her younger sister are $18$ and $12$ years.