1. solving equations

Please assist me in solving the following equation: (3t - 5) / (t - 1) = 2 + (2t) / (1 - t) Would t= -1 or 1 ? Thanks!

2. Originally Posted by jay1
Please assist me in solving the following equation: (3t - 5) / (t - 1) = 2 + (2t) / (1 - t) Would t= -1 or 1 ? Thanks!
You should be able to see straight away that $t \neq 1$ because it would create a $0$ denominator.

$\frac{3t - 5}{t - 1} = 2 + \frac{2t}{1 - t}$

$\frac{3t - 5}{t - 1} = 2 - \frac{2t}{t - 1}$

$\frac{3t - 5}{t - 1} + \frac{2t}{t - 1} = 2$

$\frac{5t - 5}{t - 1} = 2$

$5t - 5 = 2(t - 1)$

$5t - 5 = 2t - 2$

$3t = 3$

$t = 1$.

So what this means is that this function will be almost identical to a different function - the only difference being that there is a point of discontinuity at $t = 1$. Because of this point of discontinuity, this equation does not have a solution.

3. Originally Posted by Prove It
You should be able to see straight away that $t \neq 1$ because it would create a $0$ denominator.

$\frac{3t - 5}{t - 1} = 2 + \frac{2t}{1 - t}$

$\frac{3t - 5}{t - 1} = 2 - \frac{2t}{t - 1}$

$\frac{3t - 5}{t - 1} + \frac{2t}{t - 1} = 2$

$\frac{5t - 5}{t - 1} = 2$

$5t - 5 = 2(t - 1)$

Okay, that's what was confusing me.. Thanks!

$5t - 5 = 2t - 2$

$3t = 3$

$t = 1$.

So what this means is that this function will be almost identical to a different function - the only difference being that there is a point of discontinuity at $t = 1$. Because of this point of discontinuity, this equation does not have a solution.

Okay, that's what was confusing me.. Thanks!