Two snails climbed up a tree at a constant rate. A person measured and recorded their respective distances above the ground. • Snail A was 12.5 inches above the ground at 10 minutes and 16 inches above the ground at 24 minutes. • Snail B started at 3 inches above the ground and climbed 0.3 inch per minute. The snails continued at the same speeds. • Determine the amount of time, in minutes, it took for the two snails to be the same distance above the ground. • Include an equation to represent each snail's distance above the ground, y, in terms of x, the minutes elapsed since the measurement started. Show your work or explain your answer.

Using m = (y₂ - y₁)/(x₂ - x₁) , we find the rate at which Snail A moves which is also the gradient of the line with which it moves. Given that Snail A was 12.5 inches above the ground at 10 minutes and 16 inches above the ground at 24 minutes, x₁ = 10 min, y₁ = 12.5 inches, x₂ = 24 min and y₂ = 16 inches.

So, m = (y₂ - y₁)/(x₂ - x₁)

m = (16 - 12.5)/(24 - 10) = 3.5/14 = 0.25

Also, the equation of the line,

(y - y₁)/(x - x₁) = m

(y - 12.5)/(x - 10) = 0.25

cross-multiplying, we have

y - 12.5 = 0.25(x - 10)

expanding the bracket, we have

y - 12.5 = 0.25x - 2.5

adding 12.5 to both sides, we have

y - 12.5 + 12.5 = 0.25x - 2.5 + 12.5

y = 0.25x + 10

Since Snail B started at 3 inches above the ground and climbed 0.3 inch per minute. The snails continued at the same speeds, its distance y moved is y = 0.3t + 3

We now find the time x it takes the snails to cover the same distance by equating both expressions. So,

0.25x + 10 = 0.3x + 3

collecting like terms, we have

10 - 3 = 0.3x - 0.25x

7 = 0.05x

0.05x = 7

x = 7/0.05

x = 140 minutes