Question 146545
Let old road speed be s, new road speed be s+5; d=160+90, t=6
d=ts
d/s=t
160/s + 90/(s+5)=6
160(s+5)+90s=6s(s+5) Multiply each side by the LCM to eliminate fractions.
160s+800+90s=6s^2+30s
6s^2+30s-160s-90s-800=0 Subtract the left side from the right.
6s^2-220s-800=0
2(3s^2-110s-400)=0
(s-40)(3s+10)=0 Factor (or use the quadratic formula)
s=40
s+5=45 mph answer
.
Ed