What value of 't' makes the expressions 6t, t+5, and 2t+4 equal?

Question

Why can't I do this?

6t = t + 5 = 2t + 4

6t - t - 5 - 2t - 4 = 0

3t - 9 = 0

t = 3

Please explain!

However, when I do this, (which I know to be correct) it works:

t + 5 = 2t + 4

t = 2t - 1

-t = -1

t = 1

1. IT is correct!

Why can't I take 3 expressions altogether and solve it, while I can take 2 expressions at a time and solve the third.

You lost me on the third step. [0 = 5(1-t) = 4(1-t)], how did you simply get rid of '4' & '5'.

12:02 AM

Since the result of the multiplication in both cases was zero, this could only be possible if 1-t is zero.

12:23 AM

But, it could also be reached stepwise:

0=5(1-t)=4(1-t) Then divide all 3 by 5

0/5=(5/5)(1-t)=(4/5)(1-t) Giving

0=1-t=(4/5)(1/t)

12:26 AM

Typo on last, should be 0=1-t=(4/5)(1-t)

Okay, that would mean after the last step, you could also factor out (1-t) from all 3 sides as well.

0/(1-t) = (1-t)/(1-t) = (4/5)[(1-t)/(1-t)]

0 = 1 = (4/5) Not Correct

Also, If you also try to cancel out (4/5) from --> 0=1-t=(4/5)(1-t), it would simply go to

0/(5/4) = (1-t)(5/4) = 1-t

0 = (5/4)(1-t) = 1-t

Something is not right.

Answer 1

You make the error when trying to transform all 3 equations at once. When you try to eliminate one of the three, you have to add in the inverse to BOTH remaining equations, not just one. So,

6t = t + 5 = 2t + 4 -> 0 = 5 - 5t = 4 - 4t -> 0 = 5(1-t) = 4(1-t)

Thus t=1