I have a hard math question to ask :(?

d = r*t where "r" is rate of speed per hour and w= wind in mph 3000= (r - w)*6 and 3000= (r + w)* 5 TakeEQ#1 and ÷ all by 6 and EQ#2 and ÷ all by 5 r - w = 500 r + w= 600......add vertically ----------------- 2r + 0 = 1100 r=1100/2 r= 550 mph

You have two variables. X is the rate of speed of the plane, W is the rate of speed of the wind. What you did wrong: The rate of the plane minus the rate of the wind was 3000/6, not the rate of the wind alone. Break down into simpler wording. The question is basically: The rate of the plane against the push from the wind = 3000km/6hours, or 500km/h The rate of the plane plus the push of the wind = 3000km/5hours, or 600km/h X=km/hr of the plane, W=km/hr of the wind. Put into equations: So x-w=500 and x+w=600 Take one equation and represent w. w=600-x Now the two equations look like this: W=600-X and X-W=500 Then substitute in for w in the second equation so you are only solvinf for one varaible, so: (I'll show all minor steps) x-(600-x)=500 x-600 + x = 500 2x-600=500 2x-600+600=500+600 2x=1100 2x/2=1100/2 x=550km/hr so the plane in still air would be going 550km/hr then solve for w where x=550 550-w=500 550+w=600 W=600-550, W=550-500 thus W=50km/hr So the wind is 50kh/hr. Hope that was clear. (looks like I learned a different method than the others, but it has been many years since I was in school LOL - and the method I describe is applicable in more complex math problems you will encounter later in life)

You need to use both times to calculate the speed in both directions. The effective speed into the wind is the plane's speed minus the wind speed, x - w. The effective speed with the wind is the plane's speed plus the wind speed, x + w. These are the two equations: x - w = 3000 / 6 = 500 [upwind flight covers 3000km in 6 hours (at 500km per hour)] x + w = 3000 / 5 = 600 [downwind flight covers 3000km in 5 hours (at 600km per hour)] x - w = 500 x + w = 600 Add the two equations to eliminate w: x - w = 500 +(x + w = 600) --------------------- 2x = 1100 x = 550 The plane flies at 550km per hour and the wind is blowing at 50km per hour.

You have two variables. X is the rate of speed of the plane, W is the rate of speed of the wind. What you did wrong: The rate of the plane minus the rate of the wind was 3000/6, not the rate of the wind alone. Break down into simpler wording. The question is basically: The rate of the plane against the push from the wind = 3000km/6hours, or 500km/h The rate of the plane plus the push of the wind = 3000km/5hours, or 600km/h X=km/hr of the plane, W=km/hr of the wind. Put into equations: So x-w=500 and x+w=600 Take one equation and represent w. w=600-x Now the two equations look like this: W=600-X and X-W=500 Then substitute in for w in the second equation so you are only solvinf for one varaible, so: (I'll show all minor steps) x-(600-x)=500 x-600 + x = 500 2x-600=500 2x-600+600=500+600 2x=1100 2x/2=1100/2 x=550km/hr so the plane in still air would be going 550km/hr then solve for w where x=550 550-w=500 550+w=600 W=600-550, W=550-500 thus W=50km/hr So the wind is 50kh/hr. Hope that was clear. (looks like I learned a different method than the others, but it has been many years since I was in school LOL - and the method I describe is applicable in more complex math problems you will encounter later in life)

You need to use both times to calculate the speed in both directions. The effective speed into the wind is the plane's speed minus the wind speed, x - w. The effective speed with the wind is the plane's speed plus the wind speed, x + w. These are the two equations: x - w = 3000 / 6 = 500 [upwind flight covers 3000km in 6 hours (at 500km per hour)] x + w = 3000 / 5 = 600 [downwind flight covers 3000km in 5 hours (at 600km per hour)] x - w = 500 x + w = 600 Add the two equations to eliminate w: x - w = 500 +(x + w = 600) --------------------- 2x = 1100 x = 550 The plane flies at 550km per hour and the wind is blowing at 50km per hour.