Given u=(t-3)^-1; v=3(t-6)^-1; z=(t^2+t)^2 divided by t^-1, solve u+v=z.?

Question

Please explain every step by step how you solved the equation because our teacher makes us write a paragraph after we have solved the equation about how we solved it. Thank you verymuch!

Answer 1

u+v=z
(t-3)^-1 + 3(t-6)^-1 = (t^2+t)^2 divided by t^-1
(t-3)^-1 + 3(t-6)^-1 = (t^2+t)^2 / t^-1
(t-3)^-1 + 3(t-6)^-1 = t (t^2+t)^2
multiply by (t-3)(t-6):
(t-6) +3 (t-3) = t (t^2+t)^2 (t-3)(t-6)

Answer 2

Assuming you made a typo in z,

u = (t - 3)^(-1)
v = 3(t - 6)^(-1)
z = [(t^2 + t)^[-1]]/t^(-1)

Solve u + v = z.

The first thing you must do is eliminate all negative exponents. Note that 1/x^a is the same as x^(-a), i.e. every time we move through the dividing line, our exponent changes sign.

Also, note that z can be reduced.

u = 1/(t - 3)^1 = 1/(t - 3)
v = 3/(t - 6)
z = t/(t^2 + t) =t/[t(t+1)] = 1/(t+1)

Now, we substitute u, v, and z, for u + v = z

1/(t - 3) + 3/(t - 6) = 1/(t+1)

Eliminate all fractions by multiplying both sides by (t - 3)(t - 6)(t + 1). Not only will all fractions be eliminated, but stuff will be left behind.

(t - 6)(t + 1) + (t - 3)(t + 1) = [t - 3] [t - 6]

Notice that on the left hand side, each term has a (t+1). We can actually group them into one.

(t + 1) [ (t - 6) + (t - 3) ] = [t - 3] [t - 6]

Now, let's expand the second set of brackets.

(t + 1) (2t - 9) = (t - 3) (t - 6)

And now let's expand everything.

2t^2 - 9t + 2t - 9 = t^2 - 6t - 3t + 18

Combine like terms,

2t^2 - 7t - 9 = t^2 - 9t + 18

Now, bring everything over to the left hand side, giving us a quadratic equation.

t^2 + 2t + 9 = 0

We plug this into the quadratic equation now.

t = [-b +/- sqrt(b^2 - 4ac)]/2a
t = [-2 +/- sqrt(4 - 4(1)(9))]/2

At this point, we can already see that there is no solution, since we'll be taking sqrt (4 - 4(1)(9)) = sqrt (4 - 36) = sqrt (-32), which doesn't have any real solutions.