SOLUTION: Brian and Jake left their homes, which are 500 miles apart, and drove straight toward each other. It took 4 hours for the two to meet. If Jake's speed was 15 mph slower than Brian'

Question 147094: Brian and Jake left their homes, which are 500 miles apart, and drove straight toward each other. It took 4 hours for the two to meet. If Jake's speed was 15 mph slower than Brian's speed, what was Brian's speed?
Found 2 solutions by checkley77, nerdybill:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
4X+4(X-15)=500
4X+4X-60=500
8X=500+60
8X=560
X=560/8
X=70 MPH FOR THE BRIAN'S SPEED.
70-15=55 FOR JEFF'S SPEED
PROOF:
4*70+4*55=500
280+220=500
500=500

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Brian and Jake left their homes, which are 500 miles apart, and drove straight toward each other. It took 4 hours for the two to meet. If Jake's speed was 15 mph slower than Brian's speed, what was Brian's speed?
.
Use D = RT
where
D = distance
R = rate (mph)
T = time (hours)
.
Let x = Brian's speed
then
x-3 = Jake's speed
.
The idea is that the combined "distance" traveled by Brian and Jake should be 500 miles.
distance Brian traveled = RT = 4x
distance Jake traveled = RT = 4(x-3)
.
"distance Brian traveled" + "distance Jake traveled" = 500
4x + 4(x-3) = 500
4x + 4x - 12 = 500
8x = 512
x = 64 mph (Brian's speed)
x-3 = 64-3 = 61 mph (Jake's speed)

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