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Chapter 1: Rational Numbers > Distributivity of multiplication over addition

Distributivity of multiplication over addition

To understand this, consider the rational numbers −34 , 23 and −56.

Check the distributivity of multiplication over addition for −34 , 23 and −56.

Instructions

Check the distributive property

  • Evaluating: −34×23+−56, first add the numbers in the bracket. The value of the addition is by taking the LCM i.e. .
  • Now multiply both the numbers. We get the result: which can be simplified to
  • This same value can be achieved if we multiply −34 to all the terms within the bracket. Let's verify it.
  • Evaluating −34×23 we get: (Enter simplified number)
  • Now evaluating −34×−56 we get: (Enter simplified number)
  • Now adding both the number.
  • We get the value:
  • We have found that both the values are equal.
  • Thus, we have proved the following result.

Distributivity of Multiplication over Addition and Subtraction

For all rational numbers a, b and c,

a (b + c) = a b a c

a (b – c) = a b a c

Find using distributivity:

(i) 75×−312+75×512

Instructions

To solve this using distributive property, we can factor out the common term 7/5.
As 75×−312+75×512=75×−312+512.
First, let's simplify the expression inside the parentheses.
−312+512 = −3+512 = = .
Now, multiply the common factor 75 by the result : 75 × 16 = 730.
So, the result of distributive property = 730.

(ii) 916×412+916×−39

Instructions

To solve this using distributive property. Lets proceed step by step.
916 × 412 ;(simplify) 412 = .
Now, calculate 916 × 13 = 948 = .
2.916 × −39 ;(simplify) −39 = -().
Now, calculate 916 × −13 = −948 = -().
Now, add the results from step1 and step2 = 316 + −316 = .
Therefore the value is 0.

Example 3: Evaluate the expression: 25×−37−114−37×35

Instructions

25×−37−114−37×35

  • We can adjust the terms using property.
  • We can also re-write the expression by replacing the subtraction sign with addition.
  • We can now take the common number from the first two brackets and keeping the remaining terms in a bracket.
  • The value within the bracket is .
  • Solving we get: which on further simplification becomes .
  • Hence, we found the answer.