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John and Jane go rock-climbing together. John climbs a height of (x+5) miles in (x-1) hours and Jane climbs a height of (x+11) miles in (x+1) hours. Is it possible that they were climbing at the same speed? If so, what must have been their speed, in miles per hour? If not, use 0 as your answer.

Jan 16, 2019

#1
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If climbing at same speed, their rates will be equal   (if they start at the same height)

John rate = (x+5) / (x-1)          = Jane rate (x+11) / (x+1)

(x+5)(x+1) = (x+11)(x-1)

x^2 + 6x + 5 = x^2 +10x-11

6x + 5 = 10x -11

16 =4x

x=4                John speed = (4+5)/(4-1) = 3 mph      Jane speed   (4+11)/(4+1) = 3 mph

Jan 17, 2019

#1
+1

If climbing at same speed, their rates will be equal   (if they start at the same height)

John rate = (x+5) / (x-1)          = Jane rate (x+11) / (x+1)

(x+5)(x+1) = (x+11)(x-1)

x^2 + 6x + 5 = x^2 +10x-11

6x + 5 = 10x -11

16 =4x

x=4                John speed = (4+5)/(4-1) = 3 mph      Jane speed   (4+11)/(4+1) = 3 mph

ElectricPavlov Jan 17, 2019