Subtraction of Algebraic Expressions

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 + \(\frac{5}{4}\)