From the sum of 3a−2b+4c and 3b−2c subtract a−b−c.
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Solution
The value of terms as per the question is calculated as shown below (3a−2b+4c)+(3b−2c)−(a−b−c) =3a−2b+4c+3b−2c−a+b+c On further calculation, we get =3a−a+3b+b−2b+4c+c−2c =2a+2b+3c Hence, (3a−2b+4c)+(3b−2c)−(a−b−c)=2a+2b+3c