Linear Word problems

Introduction An equation which involves one or more variables whose power is 1 is called a linear equation. Example: 2x = 8 5x – 2 = 13

The sum of two consecutive number The sum of two consecutive odd number Numerical Expression 1 + 3 3 + 5 5 + 7 Algebraic expression (x) + (x + 2) Numerical Expression 1 + 2 2 + 3 3 + 4 Algebraic expression (x) + (x+1) The sum of two consecutive even number Numerical Expression 2 + 4 4 + 6 6 + 8 Algebraic expression (x) + (x + 2)

consecutive positive odd Example 1: Find two consecutive odd integers whose sum is 32. Solution: The algebraic expression for sum of consecutive odd integers (x) + (x + 2) Then (x) + (x + 2) = 32 2x + 2 = 32 (Adding all x’s) 2x + 2 – 2 = 32 -2 (Subtracting 2 on both sides) 2x = 30 Given that, Sum of two consecutive positive odd integers =32 x = 15 = First odd number. Next consecutive odd number = x + 2 = 15 + 2 = 17 (Substituting value of x = 15) Ans:- The sum of two consecutive odd integers be 15 and 17.

Therefore an obtained fraction = Example 2: The denominator of a fraction is 2 more than its numerator. If one is added to both the numerator and their denominator the fraction becomes . Find the fraction. Solution: Let the numerator be x. Then denominator is x + 2 Therefore an obtained fraction = If one added to both the numerator and denominator then the fraction becomes Given that, denominator of a fraction is 2 more than its numerator Given that, If one is added to both the numerator and their denominator the fraction becomes = Then, 3 (x + 1) = 2 (x + 3) 3x + 3 = 2x + 6 Applying distributive law Cross multiplying both fractions

Cont……. 3x + 3 = 2x + 6 3x + 3 – 2x = 2x - 2x + 6 (Subtracting 2x on both the sides) 3x -2x + 3 = 2x -2x + 6 (Combining all x terms together) x + 3 = 6 x + 3 – 3 = 6 – 3 (Subtracting 3 on both the sides) x = 3 = Numerator. Denominator = x + 2 = 3 + 2 = 5 (Substituting value of x = 3) Ans:- The required fraction =

Given that, Arun is now half as old as his father Example 3: Arun is now half as old as his father. Twelve years ago the father’s age was three times as old as Arun. Find their present age. Solution: Arun’s age = x years Then father’s age = 2x years 12 years ago, Arun’s age = x – 12 years and his father’s age = 2x – 12 years Then 2x – 12 = 3 (x -12) 2x – 12 = 3x – 36 (Apply distributive law for 3(x-12) 2x – 12 + 12 = 3x – 36 + 12 (Add 12 on both the sides) 2x = 3x – 24 2x – 3x = 3x – 24 – 3x (Subtract 3x on both the sides) -x = - 24 x = 24. Arun’s Present age = x years = 24 years His father’s present age = 2x = 2 x 24 = 48 years. (Substituting value of x = 24) Given that, Arun is now half as old as his father Given that, Twelve years ago Father’s age was three times as old as Arun’s Ans:- Arun’s Present age = 24 years His father’s present age = 48 years.

Ans: The two supplementary angles are 66° and 114° Example 4: Among the two supplementary angles, the measure of the larger angle is 48° more than the measure of smaller. Find their measures. Solution: Let the smaller angle = x° Then, Larger angle = (x + 48)° Therefore x° + (x + 48)° = 180° (Given supplementary angle) x° + x° + 48° = 180° 2x° + 48° = 180° (Adding all x’s terms) 2x° + 48° - 48° = 180° - 48° (Subtracting 48° on both the sides) 2x° = 132° (Dividing 2 on both sides) x° = 66° = Smaller angle Larger angle = x + 48 = 66 + 48 = 114° (Substituting value of x = 66) Ans: The two supplementary angles are 66° and 114° Given That, The measure of the larger angle is 48° more than the measure of smaller.

Try these Find the two consecutive positive odd integers whose sum is 64. Arjun is 6 years older than her younger sister. After 10 years, the sum of their age will be 50 years. Find their present ages.