# Thread: Identity or No Solution

1. ## Identity or No Solution

And how to work it out?

2+1/3t = 1+1/4t

2. ## Re: Identity or No Solution

Suggestion: Multiply both sides by $12t$.

3. ## Re: Identity or No Solution

Originally Posted by ingyaningya
And how to work it out?

2+1/3t = 1+1/4t
If we try t=1 we see that it doesn't work: $2 + \frac{1}{3} \neq 1 + \frac{1}{4}$ so it's not an identity

4. ## Re: Identity or No Solution

Original Problem:
$2+\frac1{3t} = 1+\frac1{4t}$

Find common denominators:

$\frac{2*(3t)+1}{3t} = \frac{4t+1}{4t}$

$\frac{6t+1}{3t} = \frac{4t+1}{4t}$

Clear fractions:

$(3t)(4t)(\frac{6t+1}{3t} = \frac{4t+1}{4t})$

$(4t)(6t+1) = (3t)(4t+1)$

Distribute:

$24t^2+4t=12t^2+3t$

$12t^2-t=0$

Factor:
$t(12t-1) = 0$

solve for t:
t = 0 or 12t-1=0

$t=0$ or $t=\frac{1}{12}$

5. ## Re: Identity or No Solution

Originally Posted by mdhafn
Original Problem:
$2+\frac1{3t} = 1+\frac1{4t}$

Find common denominators:

$\frac{2*(3t)+1}{3t} = \frac{4t+1}{4t}$

$\frac{6t+1}{3t} = \frac{4t+1}{4t}$

Clear fractions:

$(3t)(4t)(\frac{6t+1}{3t} = \frac{4t+1}{4t})$

$(4t)(6t+1) = (3t)(4t+1)$

Distribute:

$24t^2+4t=12t^2+3t$

$12t^2-t=0$

Factor:
$t(12t-1) = 0$

solve for t:
t = 0 or 12t-1=0

$t=0$ or $t=\frac{1}{12}$
0 is not a solution since you'd be dividing by 0 in the original equation.

@OP: For the equation to be an identity all values of t must work. For no solution then there must be no value of t.

6. ## Re: Identity or No Solution

$24t^2+4t=12t^2+3t$

img.top { vertical-align: 15%; } $12t^2-t=0$

Subtracting 3t leaves +t on the left.

12t^2 + t = 0

t(12t+1)=0

t = 0 or -1/12

Drop 0 because the original equation has t in the denominator.