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Jack walked up a hill at a speed of \((x^2-11x-22)\) miles per hour. Meanwhile, Jill walked a total distance of \((x^2-3x-54)\) miles in \((x+6)\) hours. If Jack and Jill walked at the same speed, what is that speed, in miles per hour?

Logic Sep 14, 2018

#1**+1 **

Distance = Speed x Time

Solve for x:

(x + 6) (x^2 - 11 x - 22) = x^2 - 3 x - 54

Expand out terms of the left hand side:

x^3 - 5 x^2 - 88 x - 132 = x^2 - 3 x - 54

Subtract x^2 - 3 x - 54 from both sides:

x^3 - 6 x^2 - 85 x - 78 = 0

The left hand side factors into a product with three terms:

(x - 13) (x + 1) (x + 6) = 0

Split into three equations:

x - 13 = 0 or x + 1 = 0 or x + 6 = 0

Add 13 to both sides:

x = 13 or x + 1 = 0 or x + 6 = 0

Subtract 1 from both sides:

x = 13 or x = -1 or x + 6 = 0

Subtract 6 from both sides:

** x = 13 sub this in Jack's speed:(13^2 - (11*13) - 22) = 4 miles per hour.**

**Sub 4 into Jill's distance: (13^2 - (3*13) - 54) =76 miles. Divide this by her time:**

**76 / (13+6) =76/19 =4 miles per hour - Jill's speed = Jack's speed.**

Guest Sep 15, 2018

#2**+1 **

Distance / Rate = Speed...so

Jack's speed = Jill's speed

x^2 -11x - 22 = (x^2 - 3x - 54) / (x + 6)

x^2 - 11x - 22 = (x - 9)(x + 6) / (x + 6)

x^2 - 11x - 22 = x - 9 rearrange as

x^2 - 12x - 13 = 0 factor

(x - 13) ( x + 1) = 0

Set each factor to 0 and solve for x and we get that x = 13 or x = -1

Reject the second answer since itleads to a negative speed

If x = 13, the speed is 4 mph

CPhill Sep 15, 2018