Question

I don't know how to start this one.

(t/t-4)=(t+4/6)-1

Answer 1

@Precal

Answer 2

\[(\frac{ t }{ t-4 }) = (\frac{ t+4 }{ 6 }) -1\]
can you solve proportions?

Answer 3

do you think that is how you would go about solving it? or what did you learn in school about doing this?

Answer 4

you actually can simplify these terms like
t/t = 1

Answer 5

No I can not solve proportions

Answer 6

so for the first one you would have
-1/4 = ... can you do the other one?

Answer 7

-1/4 = (t+2/3) -1
lets multiply both sides by -1 to get rid of it
1/4 =-(t+2/3)

Answer 8

Where did you get t/t

Answer 9

t/t-4
t/t is 1

Answer 10

\[(\frac{ t }{ t-4 }) = (\frac{ t+4 }{ 6 }) -6 \]

\[(\frac{ t }{ t-4 }) = \left( \frac{ t+4-1 }{ 6 } \right)\]

\[(\frac{ t }{ t-4 }) - \left( \frac{ t+4-1 }{ 6 } \right) \]

Answer 11

-2/3 both sides you get
3/12 - 8/12
-5/12 = -t
t= 5/12

Answer 12

\[\frac{ \left( 6 \times t \right) - \left( t-4 \right)\left( t+3 \right) }{ 6\left( t-4 \right) }\]

Answer 13

I dont even know if that is right cuz mites is doing something diff. i would guess he is right cuz i've only done Algebra I. But i just tried to help you from what i know.

Answer 14

r.h.s =0

\[6t - \left( t^2 - t - 12 \right) = 0\]

Answer 15

I am sorry the actual problem has no parenthacies. I don't not know how to write it

Answer 16

\[6t - t^2 +t +12 = 0\]

\[t^2 - 7t - 12 = 0\]

\[\left( t-4 \right)\]

Answer 17

sry
\[\left( t-4 \right)\left( t-3 \right)\]

Answer 18

therefore t=4 or t =3

Answer 19

\[(\frac{ t }{ t-4 }) = (\frac{ t+4 }{ 6 }) - \frac{ 6 }{ 6 }=0\]

\[(\frac{ t }{ t-4 }) = \left( \frac{ t+4-6 }{ 6 } \right) = 0 \]

\[(\frac{ t }{ t-4 }) - \left( \frac{ t+4-6 }{ 6 } \right) = 0\]

\[\frac{ \left( 6 \times t \right) - \left( t-4 \right)\left( t-2 \right) }{ 6\left( t-4 \right) }=0\]
\[6t - \left( t^2 - 6t +8 \right) = 0\]

now you can solve