SOLUTION: Please help me this word problem: Kevin drove 320 miles to a mountain resort. His return trip took 20 minutes longer because his speed returning was 4 mi/h slower than his speed go

Question 214890: Please help me this word problem: Kevin drove 320 miles to a mountain resort. His return trip took 20 minutes longer because his speed returning was 4 mi/h slower than his speed going.

Found 2 solutions by RAY100, checkley77:
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
Velocity = Distance / Time,,,,or T = D/V
.
(1) T1 = 320 / V1
.
(2) T2 = 320 / V2,,,,,but T2 = T1 + 20/60 (hrs) ,,,and V2 = V1 -4
.
subst in (2)
.
{ T1 + 20/60} = 320 / {V1-4},,,and T1 = 320 /V!
.
{ (320/V1) +20/60 } = 320 /{V1-4)
.
Multiply thru by (3) V1(V1-4)
.
(320 (3) (V1-4) + 1(V1)(V1-4) = 320 (3)(V1)
.
960 V1 - 3840 + V1^2 -4V1 = 960V1
.
v1^2 -4V1 -3840 =0,,,,,factor
.
(V1 -64)(V1+60) =0
.
V1= 64, -60 ( not practical)
.
check,,,V1= 64,,,V2 = 60
.
T1 = 320 /64 = 5 hrs
.
T2 = 5 hrs + 1/3 hr = 5.33 hrs
.
T2 = 320 / 60 = 5.333,,,,,ok
.

Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
D=RT
320=RT OR R=320/T
320=(R-4)(T+1/3)
320=(320/T-4)(T+1/3)
320=320-4T+320/3T-4/3
-4T+320/3T-4/3=0
(-4T*3T+320-4T)/3T=0
(-12T^2+320-4T)/3T=0
12T^2+4T-320=0
4(3T^2+T-80)=0

T=(-1+-SQRT[1^1-4*3*-80])/2*3
T=(-1+-SQRT[1+960])/6
T=(-1+-SQRT961)/6
T=(-1+-31)/6
T=-32/3=-5.333
T=(-1+31)/6
T=30/6
T=5 HRS. GOING.
PROOF;
320=R*5
R=320/5
R=64 MPH. GOING
320=(64-4)(5+1/3)
320=60*16/3
320=960/3
960=960

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