# Susan is half the age of Amy. Five years ago, Susan was three-fourths as old as Amy. How old is each?

Aug 8, 2017

See a solution process below:

#### Explanation:

Let's call Susan's currrent age: $s$

And, let's call Amy's current age: $a$

Right now we can say:

$s = \frac{1}{2} a$

But, fives years ago when Susan was $\left(s - 5\right)$ years old and Amy was $a - 5$ years old we can write:

$\left(s - 5\right) = \frac{3}{4} \left(a - 5\right)$

From the first equation we can substitute $\frac{1}{2} a$ for $s$ into the second equation and solve for $a$:

$\left(s - 5\right) = \frac{3}{4} \left(a - 5\right)$ becomes:

$\left(\frac{1}{2} a - 5\right) = \frac{3}{4} \left(a - 5\right)$

$\frac{1}{2} a - 5 = \left(\frac{3}{4} \times a\right) - \left(\frac{3}{4} \times 5\right)$

$\frac{1}{2} a - 5 = \frac{3}{4} a - \frac{15}{4}$

$- \textcolor{red}{\frac{1}{2} a} + \frac{1}{2} a - 5 + \textcolor{b l u e}{\frac{15}{4}} = - \textcolor{red}{\frac{1}{2} a} + \frac{3}{4} a - \frac{15}{4} + \textcolor{b l u e}{\frac{15}{4}}$

$0 - 5 + \textcolor{b l u e}{\frac{15}{4}} = - \textcolor{red}{\frac{1}{2} a} + \frac{3}{4} a - 0$

$- 5 + \textcolor{b l u e}{\frac{15}{4}} = - \textcolor{red}{\frac{1}{2} a} + \frac{3}{4} a$

$\left(\frac{4}{4} \times - 5\right) + \textcolor{b l u e}{\frac{15}{4}} = \left(\frac{2}{2} \times - \textcolor{red}{\frac{1}{2} a}\right) + \frac{3}{4} a$

$- \frac{20}{4} + \textcolor{b l u e}{\frac{15}{4}} = - \textcolor{red}{\frac{2}{4} a} + \frac{3}{4} a$

$- \frac{5}{4} = \left(- \textcolor{red}{\frac{2}{4}} + \frac{3}{4}\right) a$

$- \frac{5}{4} = \frac{1}{4} a$

$\textcolor{red}{4} \times - \frac{5}{4} = \textcolor{red}{4} \times \frac{1}{4} a$

$\cancel{\textcolor{red}{4}} \times - \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} = \cancel{\textcolor{red}{4}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} a$

$- 5 = a$

$a = - 5$

We can now substitute $- 5$ for $a$ in the first equation and calculate Susan's age.

$s = \frac{1}{2} a$ becomes:

$s = \left(\frac{1}{2} \times - 5\right)$

$s = - \frac{5}{2}$

$s = - 2.5$

This problem does not make sense unless you state:

Amy is not born yet and will be born in 5 years

Susan is also not born yet but will be born in $2 \frac{1}{2}$ years.