Solve the following:
$\frac{y-1}{3}-\frac{y-2}{4}=1$

Given :

The given expression is $\frac{y-1}{3}-\frac{y-2}{4}=1$.

To do :

We have to find the value of y.

Solution :

 $\frac{y-1}{3}-\frac{y-2}{4}=1$

LCM of 3 and 4 is 12.

So,

$\frac{4(y-1)}{3\times 4}-\frac{3(y-2)}{4\times 3}=1$

$\frac{4y-4}{12}-\frac{3y-6}{12}=1$

$\frac{4y-4 - (3y-6)}{12} = 1$

$\frac{4y-4 - 3y+6}{12} = 1$

$\frac{4y- 3y+6 -4}{12} = 1$

$\frac{y+2}{12} = 1$

$y+2 = 12$

$y = 12-1$

$y = 11$

Therefore, the value of y is 11.