You can
put this solution on YOUR website!Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let x=speed of plane
And let y=speed of the ship
Distance initially travelled by plane=4x
Distance initially travelled by ship=25y so:
4x+25y=1580----------------------------------eq1
and
4(x/2)+25(5y/4))=1315 (Note: 1/4 greater than y=4y/4+y/4=5y/4)- multiply each term by 4
8x+125y=5260----------------------------------eq2
multiply eq1 by 2, and we get:
8x+50y=3160
and then subtract it from eq2:
75y=2100 divide both sides by 75
y=28 mph---------------speed of ship
substitute y=28 into eq1
4x+25*28=1580
4x+700=1580 subtract 700 from each side
4x=880 divide each side by 4
x=220 mph speed of plane
CK
220*4+25*28=1580
880+700=1580
1580=1580
and
(1/2)*220*4+25*(5/4)*28=1315
440+875=1315
1315=1315
Hope this helps--ptaylor
