Math problem related to pythagorean theorem?

Let y mph be your speed.
Then, the speed of your friend = (y + 5) mph

Refer to the figure below.
O is the starting point. After 3 hours, you reach A and your friend reached B.
OA = 3y miles
OB = 3(y + 5) miles

OA² + OB² = AB²
(3y)² + [3(y + 5)]² = 256.89²
9y² + 9y² + 90y + 225 = 65992.4721
18y² + 90y - 65767.4721 = 0
y = [-90 ± √(90² + 4×18×65767.4721)] / (2×18)
y = 58
y + 5 = 63

Answer:
Your speed = 58 mph
Your friend's speed = 63 mph

Recall: s = d/t → where s is the speed, d is the distance, t is the time


You drive north for 3 hours. Adapt the previous formula to this case:

s₁ = d₁/t₁

d₁ = s₁.t₁ → given that the time is 3 hours

d₁ = 3.s₁


Your friend drives east at a speed that is 5 mph higher than you. Adapt the previous formula to this case:

s₂ = d₂/t₂

d₂ = s₂.t₂ → given that the time is 3 hours

d₂ = 3.s₂ → the speed of your friend is 5 mph higher than your → s₂ = s₁ + 5

d₂ = 3.(s₁ + 5)

d₂ = 3.s₁ + 15 → recall: d₁ = 3.s₁

d₂ = d₁ + 15


After 3 hours the distance between you and your friend is 256.89 miles:

d² = d₁² + d₂² → given that: d = 256.89

256.89² = d₁² + d₂² → recall: d₂ = d₁ + 15

256.89² = d₁² + (d₁ + 15)²

256.89² = d₁² + d₁² + 30.d₁ + 15²

2.d₁² + 30.d₁ = 256.89² - 15²

2.d₁² + 30.d₁ = 65767.4721

d₁² + 15.d₁ = 32883.73605

d₁² + 15.d₁ + 7.5² = 32883.73605 + 7.5²

(d₁ + 7.5)² = 32939.98605

d₁ + 7.5 = ± √32939.98605

d₁ = - 7.5 ± √32939.98605 → the distance is a positive value of course

d₁ = - 7.5 + √32939.98605


Recall: s₁ = d₁/t₁

s₁ = (- 7.5 + √32939.98605) / 3

s₁ = 58 mph


Recall: s₂ = s₁ + 5

s₂ = 63 mph

Speed North is x m.p.h.

so, distance after 3 hours is 3x

Speed East is x + 5 m.p.h.

so, distance after 3 hours is 3(x + 5) => 3x + 15 

Then, x² + (3x + 15)² = 256.89²

so, 10x² + 90x + 225 = 65992.4721

i.e. 10x² + 90x - 65767.4721 = 0     

Using the quadratic formula we get:

x = [-90 ± √(90² - 4(10)(-65767.4721))]/20

so, x = (-90 ± √2638798.884)/20

Then, x = 76.7 m.p.h.

:)>

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