English

Take Away: Y 3 3 + 7 3 Y 2 + 1 2 Y + 1 2 from 1 3 − 5 3 Y 2 - Mathematics

Advertisements
Advertisements

Question

Take away: 

\[\frac{y^3}{3} + \frac{7}{3} y^2 + \frac{1}{2}y + \frac{1}{2} \text { from } \frac{1}{3} - \frac{5}{3} y^2\]

Answer in Brief

Solution

The difference is given by:

\[\left( \frac{1}{3} - \frac{5}{3} y^2 \right) - \left( \frac{y^3}{3} + \frac{7 y^2}{3} + \frac{y}{2} + \frac{1}{2} \right)\]

\[ = \frac{1}{3} - \frac{5}{3} y^2 - \frac{y^3}{3} - \frac{7 y^2}{3} - \frac{y}{2} - \frac{1}{2}\]

\[= \frac{1}{3} - \frac{1}{2} - \frac{y}{2} - \frac{5}{3} y^2 - \frac{7 y^2}{3} - \frac{y^3}{3}\]   ( Collecting like terms)

= \[\left( \frac{2 - 3}{6} \right) - \frac{y}{2} + \left( \frac{- 5 - 7}{3} \right) y^2 - \frac{y^3}{3}\] 

\[= - \frac{1}{6} - \frac{y}{2} - 4 y^2 - \frac{y^3}{3}\]  (Combining like terms. )

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Algebraic Expressions and Identities - Exercise 6.2 [Page 5]

APPEARS IN

RD Sharma Mathematics [English] Class 8
Chapter 6 Algebraic Expressions and Identities
Exercise 6.2 | Q 3.4 | Page 5
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×