Question 1028613
DISCLAIMER: No guarantees that this is the way your teacher thinks the problem should be solved. (The result is correct, though).
 
ONE WAY:
{{{x}}}= distance the girl walked, in miles
{{{20-x}}}= distance the girl rode in her dad's car, in miles
{{{d/3}}}= time the girl spent walking towards school, in hours
{{{(30-d)/30}}}= time the girl spent riding in her dad's car towards school, in hours
{{{"2 hours 10 minutes"=2&1/6}}}{{{hours=13/6}}}{{{hours}}}= total time walking and riding to school
{{{d/3+(30-d)/30=13/6}}} is our equation.
Multiplying times {{{30}}} ,
{{{10d+30-d=65}}}
{{{10d-d=65-30}}}
{{{9d=45}}}
{{{d=45/9}}}
{{{highlight(d=5)}}}
So the girl walked {{{highlight(5miles)}}} before her dad picked her up.
 
USING SYSTEMS OF EQUATIONS:
{{{x}}}= time walking to school, in hours
{{{y}}}= time riding to school, in hours
{{{x+y=130/60=13/6}}}= total time walking and riding to school
{{{3x}}}= miles walked to school
{{{30y}}}= miles traveled by car going to school
{{{3x+30y=20}}}
{{{system(x+y=13/6,3x+30y=20)}}}-->{{{system(30x+30y=65,3x+30y=20)}}}-->{{{27x=45)}}}-->{{{x=5/3}}}-->{{{3x=highlight(5)}}}
So the girl walked {{{highlight(5miles)}}} before her dad picked her up.