Bob and wendy left home to walk together to a restaurant for dinner. they started out walking at a constant pace of 3 mph. at precisely the halfway point, bob realized he had forgotten to lock the front door of their home. wendy continued on to the restaurant at the same constant pace. meanwhile, bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join wendy. how long did wendy have to wait for bob at the restaurant? (1) bob’s average speed for the entire journey was 4 mph.
(2) on his journey, bob spent 32 more minutes alone than he did walking with wendy.

31.98 min

Step-by-step explanation:

if we divide the route to the restaurant in two halfs we can see that bob walked 4 halfs (the first half going to the restaurante, the first one again returning home, and one more time the first half going to the restaurant again)

if his average speed was 4mph, and we know the first quarter of the journy he was walking at 3 mph, we can say

1/4 * 3mph + 3/4 * Vmph = 4 mph

then

V = (4mph - 1/4 * 3mph) / (3/4)  mph

V= 4.33 mph the speed when bob walked alone.

if we know he spent 32 more minutes alone than he did walking whith wendy

1/4 x   / 3mph + 32min = 3/4 x / 4.33                 32min = 0.533h

x= 5.93miles (total journey for bob, 4 halfs the distanche from home to the restaurant)

then, from they got separated ways Wendy walked one half, at 3 mph

1/4 x * 3mph = Twendy  

She walked for 29.65min (0.49h)

Bob walked 3 halfs at 4.33mph

3/4 x * 4.33 mph = Tbob

he walked alone for 61.63min (1.02h)

she waited him for 31.98 min

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.