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.
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