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#### Additive inverse of an expression:

For every positive number, there exists a negative number such that their **sum is zero**. These two numbers are called the **additive inverse** of the each other.

**Example:** If we take a positive number, say, 7 then there exists (-7) such that 7 + (-7) = 0.

Here, we say that ‘-7’ is the additive inverse of 7 and 7 is the additive inverse of ‘-7’.

Similarly, for every algebraic expression there exists another algebraic expression such that their sum is zero. These two expressions are called additive inverse of each other.

#### Subtraction of Like terms:

The difference of like terms is a like term whose coefficient is the difference of the numerical coefficients of the like terms.

**Example 1:**

(4\(x^2\) – 6\(x^2\)) = (4 – 6)\(x^2\)

= -2\(x^2\)

**Example 2:**

(4xy – 6xy –2xy) = (4 – 6 – 2)xy

= –4xy.

#### Subtraction of Algebraic Expressions:

In order to subtract one algebraic expression from another, **change the sign of each term of the expression to be subtracted and then add to the other expression.**

### Subtraction of Algebraic Expressions Example 1:

Subtract (2\(x^2\) – 5x + 7) from (3\(x^2\) + 4x – 6).

**Solution:** We have, (3\(x^2\) + 4x – 6) – (2\(x^2\) – 5x + 7)

= 3\(x^2\) + 4x – 6 – 2\(x^2\) + 5x – 7

= (3-2)\(x^2\) + (4 + 5)x + (-6 – 7)

= \(x^2\) + 9x – 13

### Subtraction of Algebraic Expressions Example 2:

Take away (\(\frac{8}{5}\)\(x^2\) – \(\frac{2}{3}\)\(x^3\) + \(\frac{3}{2}\)x – 1) from (\(\frac{1}{5}\)\(x^3\) – \(\frac{3}{2}\)\(x^2\) + \(\frac{2}{3}\)x + \(\frac{1}{4}\))

**Solution:** We have, (\(\frac{1}{5}\)\(x^3\) – \(\frac{3}{2}\)\(x^2\) + \(\frac{2}{3}\)x + \(\frac{1}{4}\)) – (\(\frac{8}{5}\)\(x^2\) – \(\frac{2}{3}\)\(x^3\) + \(\frac{3}{2}\)x – 1)

= \(\frac{1}{5}\)\(x^3\) – \(\frac{3}{2}\)\(x^2\) + \(\frac{2}{3}\)x + \(\frac{1}{4}\) – \(\frac{8}{5}\)\(x^2\) + \(\frac{2}{3}\)\(x^3\) – \(\frac{3}{2}\)x + 1

= (\(\frac{1}{5}\) + \(\frac{2}{3}\))\(x^3\) + (-\(\frac{3}{2}\) – \(\frac{8}{5}\))\(x^2\) + (\(\frac{2}{3}\) – \(\frac{3}{2}\))x + (\(\frac{1}{4}\) + 1)

= \(\frac{3 + 10}{15}\)\(x^3\) + (\(\frac{-15 – 16}{10}\))\(x^2\) + \(\frac{4 – 9}{6}\)x + \(\frac{1 + 4}{4}\)

= \(\frac{13}{15}\)\(x^3\) – \(\frac{31}{10}\)\(x^2\) – \(\frac{5}{6}\)x + \(\frac{5}{4}\)