## Algebra: A Combined Approach (4th Edition)

Let the time taken by the jogger going to the park be = $x$ hours. Thus, she takes 1 hour more ($x+1$ hours) to return home from the longer route. The rate is constant, so it does not matter what value we take. So, let the rate be = $r$. Thus, while going to the park, Distance = 12 miles, time = $x$ hours. Thus Distance = Rate $\times$ Time = $r(x)$ Thus 12 = $r(x)$ $\frac{12}{r}$ = $x$ x = $\frac{12}{r}$ And while returning , Distance = 18 miles, time = $(x+1)$ hours Thus 18 = $r(x+1)$ $\frac{18}{r}$ = $x+1$ $x$ = $\frac{18}{r}-1$ = $\frac{18-r}{r}$ Thus, $x$ = $\frac{12}{r}$ = $\frac{18-r}{r}$ hours. Thus: $\frac{12}{r}$ = $\frac{18-r}{r}$ Multiplying both sides by $r$, $\frac{12}{r}$($r$) = $\frac{18-r}{r}$$(r)$ ....(Since $\frac{r}{r}=1$) Thus $12=18-r$ Thus, adding $r$ on both sides, $12+r=18-r+r$ = 18 Thus, subtracting 12 from both sides, $12+r-12$ = $18-12$ = 6 $r$ = 6 Thus $r$ = 6 miles/hour Jogging speed = 6 miles/hour.