The sum of the present ages of Erick and Henry is $55$. When Henry was same old as Erick's present age, Erick was $20$. What is Henry's present age?
I'm still thinking on it. Let me show what I thought
$$E + H = 55 \tag{1}$$
$$H -t = E \tag{2}$$
$$E -t = 20 \tag{3}$$
where $t = \text {passed time}$, $E = \text{Erick}$, $H = \text{Henry}$
I've gone too wrong, haven't I?
On
Solving for one variable then using the result to solve for the others will give you the correct results. Here I solve for t
$(1)$ $E+H=55$
$(2)$ $H-t=E$ $->$ $H=E+t$
$(3)$ $E-t=20$ $->$ $E=20+t$
Adding $(2)$ and $(3)$, will isolate H.
$H+E=E+2t+20$
$H=20+2t$
Substituting $(2)+(3)$ and $(3)$ in $(1)$, will give the value of t.
$20+t+20+2t=55$
$t=5$
Now the rest can be calculated fairly easily.
Seems good to me.
Three linear equations with three unknown variables. Should be solvable if they aren't dependent which they clearly aren't.
So try to solve them.
.... go ahead. Solve them... My solution is below but you should try solving them first.
One way: $H = E + t$ so substitute that back into the first to get $E + H = E+(E+t) = 2E + t = 55$. Add that to the last equation to get $(2E+t) +(E-t) = 55 + 20$ or $3E = 75$. So $E = 25$.
Plug that in and we get $25 + H = 55$ and $H - t = 25$ and $25 -t = 20$. From there is clear $H = 30$ and $t= 5$ and.... sure enough. Eric is $25$ and Henry is $30$ and $5$ years ago when Henry was $25$, Eric was $20$.