Question 831042
I do not see that problem as an "applying systems of linear equations word problem", but I will try to do it that way, and also show you an alternate way.
 
THE MORE DIRECT WAY:
Average speed, time, and distance are related by
{{{speed=distance/time}}} <--> {{{time*speed=distance}}} <--> {{{time=distance/speed}}}
 
Measuring time in hours, it took {{{1+45/60=1.75=7/4}}} hours for the whole round trip.
The first part, at average speed {{{r}}} would have taken {{{45/r}}} hours.
The return trip, at average speed {{{(3/4)r}}} would have taken {{{45/((3/4)r)}}} hours.
Adding up, we get one equation.
or {{{45/r+45/((3/4)r)=1.75}}}
and we can solve for {{{r}}} .
 
Solving:
{{{45/r+45/((3/4)r)=7/4}}}
{{{45/r+(45/r)(4/3)=7/4}}} 
{{{(45/r)(1+4/3)=7/4}}}
{{{(45/r)(7/3)=7/4}}}
{{{45(4/3)=r}}}
{{{highlight(r=60)}}} and {{{(3/4)r=(3/4)60=highlight(45)}}}
 
USING SYSTEMS OF LINEAR EQUATIONS
{{{x}}}= time (in hours) Julie spent driving at an average speed of {{{r}}} miles per hour, while going wherever Julie was going.
{{{y}}}= time (in hours) Julie spent driving at an average speed of {{{(3/4)r}}} miles per hour during the return trip.
Since {{{time*speed=distance}}} , for both parts of the trip we have
{{{rx=45}}} and {{{(3/4)ry=45}}} ,
and we could combine them into
{{{rx=(3/4)ry}}} --> {{{x=(3/4)y}}}
We also know that {{{x+y=7/4}}}
 
Solving:
{{{system(x=(3/4)y,x+y=1.75)}}} --> {{{system(x=(3/4)y,(3/4)y+y=1.75)}}} --> {{{system(x=(3/4)y,(7/4)y=7/4)}}} --> {{{system(x=(3/4)*1,y=1)}}} --> {{{highlight(system(x=3/4,y=1))}}}
So the average speed is {{{45/((3/4))=45*(4/3)=highlight(60)}}} on the way over there,
and {{{45/1=highlight(45)}}} for the return trip.