Question 1046349
There are typos in that question.
I am guessing the question meant states:
The distance from city A to city B is five times the distance from city B to city C.
The distance from city A to city D is two times the distance from city A to city B.
The distance between cities C to D is 10 miles.
 
A SHORT WRITE UP:
{{{AB=5BC}}}
{{{AD=2AB}}}
{{{CD=10}}}
{{{AD=AB+BC+CD}}}
{{{2AB=AB+BC+10}}}
{{{2(5BC)=5BC+BC+10}}}
{{{10BC=6BC+10}}}
{{{10BC-6BC=10BC=10}}}
{{{BC=10/4}}}
{{{BC=2.5}}}
{{{5BC=5(2.5)}}}
{{{5BC=12.5}}}
{{{AB=12.5}}}
The distance between city A and city B is {{{highlight(12.5miles)}}} .


A LONGER EXPLANATION, WITH A VARIABLE SO IT LOOKS MORE LIKE ALGEBRA:
Let us define a variable {{{x}}} .
{{{x}}}= the distance from city B to city C, in miles.
(We could call it {{{BC}}} , but {{{x}}} is shorter).
Since the distance from city A to city B is five times the distance from city B to city C,
the distance from city A to city B is {{{5x}}} .
The distance from A to C must be the sum of the two distances listed above:
{{{5x+x=6x}}} .
So far we have this picture:
{{{drawing(600,100,-1,29,-2,3,
line(-1,0,29,0),red(circle(12.5,0,0.2)),
red(circle(15,0,0.2)),red(circle(0,0,0.2)),
red(circle(25,0,0.2)),locate(-0.2,-0.2,A),
locate(12.3,-0.2,B),locate(14.8,-0.2,C),
locate(24.8,-0.2,D),locate(5.8,0,5x),
locate(13.5,0.15,x),red(arrow(7.5,1,0,01)),
red(arrow(7.5,1,15,1)),locate(7.1,1,6x),
locate(25.3,-1.2,highway),arrow(28.2,-1.7,28.2,0)
)}}}
We also know that the distance from city A to city D is two times the distance from city A to city B.
So, th2 distance from city A to city D is
{{{2*(5x)=10x}}} ,
and we know that the distance between cities C to D is 10 miles.
Now, the picture looks like this:
{{{drawing(600,100,-1,29,-2,3,
locate(25.3,-1.2,highway),arrow(28.2,-1.7,28.2,0),
line(-1,0,29,0),red(circle(12.5,0,0.2)),
red(circle(15,0,0.2)),red(circle(0,0,0.2)),
red(circle(25,0,0.2)),locate(-0.2,-0.2,A),
locate(12.3,-0.2,B),locate(14.8,-0.2,C),
locate(24.8,-0.2,D),locate(5.8,0,5x),
locate(13.5,0.15,x),red(arrow(7.5,1,0,01)),
red(arrow(7.5,1,15,1)),locate(7.1,1,6x),
locate(19.7,0,10),
red(arrow(12.5,2,0,2)),red(arrow(12.5,2,25,2)),
locate(12,2,10x)
)}}} .
Considering the distances shown by the arrows
between A and C {{{(6x)}}}, and a
between A and D {{{(10x)}}} ,
the distance between C and D must be their difference
{{{10x+6x=4x}}} .
Now, we can represent the situation as
{{{drawing(600,100,-1,29,-2,3,
locate(25.3,-1.2,highway),arrow(28.2,-1.7,28.2,0),
line(-1,0,29,0),red(circle(12.5,0,0.2)),
red(circle(15,0,0.2)),red(circle(0,0,0.2)),
red(circle(25,0,0.2)),locate(-0.2,-0.2,A),
locate(12.3,-0.2,B),locate(14.8,-0.2,C),
locate(24.8,-0.2,D),locate(5.8,0,5x),
locate(13.5,0.15,x),red(arrow(7.5,1,0,01)),
red(arrow(7.5,1,15,1)),locate(7.1,1,6x),
locate(19.7,0,10),
red(arrow(12.5,2,0,2)),red(arrow(12.5,2,25,2)),
locate(12,2,10x),
green(arrow(20,1,15,1)),green(arrow(20,1,25,1)),
locate(18,1,10x-6x=4x)
)}}}
Finally since the distance, in miles, between C and D is {{{10}}} and {{{4x}}} ,
{{{4x=10}}}
{{{x=10/4}}}
{{{x=5}}} is the distance, in miles, between B and C,
and the distance, in miles, between A and B is
{{{5x=5*2.5}}}
{{{5x=highlight(12.5)}}} .
The distance between city A and city B is {{{highlight(12.5miles)}}} .