l 4t+1 l = l 6t-1 l what is the absolute value?

Answer 1

You need to consider two possible solutions.
Either 4t + 1 = 6t - 1 or 4t + 1 = -(6t - 1).

In the first case 1 = 2t - 1, so 2 = 2t, so 1 = t

In the second case, 4t + 1 = -6t + 1, so 10t + 1 = 1, so 10t = 0, so t = 0.

You need to check to see which of the solutions actually work, because sometimes problems like this produce extraneous solutions that don't actually work in the original problem.

In the first case ABS(4*1 + 1) = ABS(5) = 5 and ABS(6*1 - 1) = ABS(5) = 5. So t = 1 is a solution.

In the second case ABS(4*0 + 1) = ABS(1) = 1 and ABS(6*0 - 1) = ABS(-1) = 1. So t = 0 is also a solution.