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Solve : (3t-2)/4 - (2t+3)/3 = 2/3 - t...

Solve : `(3t-2)/4 - (2t+3)/3 = 2/3 - t`

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`(3t-2)/4 - (2t+3)/3 = 2/3 - t`
Now by taking L.C.M for `4` and `3` is `12`
`(3(3t−2)−4(2t+3))/12=2/3−t`
By transposing the above equation we can write as
`=>(9t−6−8t−12)/12=2/3−t`
`=>t+12t=8+18`
Again by transposing
`t = 26/13`
`t = 2`
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