Find the greatest value of t such that \(\frac{t^2 - t -56}{t-8} = \frac{3}{t+5}.\)

Guest Jan 15, 2019

#1**+1 **

Solve for t:

(t^2 - t - 56)/(t - 8) = 3/(t + 5)

Cross multiply:

(t + 5) (t^2 - t - 56) = 3 (t - 8)

Expand out terms of the left hand side:

t^3 + 4 t^2 - 61 t - 280 = 3 (t - 8)

Expand out terms of the right hand side:

t^3 + 4 t^2 - 61 t - 280 = 3 t - 24

Subtract 3 t - 24 from both sides:

t^3 + 4 t^2 - 64 t - 256 = 0

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

(t - 8) (t + 4) (t + 8) = 0

Split into three equations:

t - 8 = 0 or t + 4 = 0 or t + 8 = 0

Add 8 to both sides:

t = 8 or t + 4 = 0 or t + 8 = 0

Subtract 4 from both sides:

t = 8 or t = -4 or t + 8 = 0

Subtract 8 from both sides:

t = 8 or t = -4 or t = -8

(t^2 - t - 56)/(t - 8) ⇒ (-56 - -8 + (-8)^2)/(-8 - 8) = -1

3/(t + 5) ⇒ 3/(5 - 8) = -1:

So this solution is correct

(t^2 - t - 56)/(t - 8) ⇒ (-56 - -4 + (-4)^2)/(-8 - 4) = 3

3/(t + 5) ⇒ 3/(5 - 4) = 3:

So this solution is correct

(t^2 - t - 56)/(t - 8) ⇒ (-56 - 8 + 8^2)/(8 - 8) = (undefined)

3/(t + 5) ⇒ 3/(5 + 8) = 3/13:

So this solution is incorrect

The solutions are:

**t = -4 or t = -8**

Guest Jan 15, 2019

#2**0 **

Factor the right numerator to get:

(t-8)(t+7) / (t-8) = 3/(t+5) the term (t-8) cancles out on the right but REMEMBER t cannot equal 8 (would make zero denominator)

t+7 = 3/ (t+5)

(t+7)*(t+5) = 3

t^2 + 12t + 35 = 3

t^2 + 12t + 32 = 0

(t+8)(t+4) = 0 Shows t = -8 or -4 the greatest of which is -4

ElectricPavlov Jan 15, 2019