1. Linear Word problems

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

3. 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)

4. 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 = (Adding all x’s)
2x + 2 – 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 = = 17 (Substituting value of x = 15)
Ans:- The sum of two consecutive odd integers be 15 and 17.

5. 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 = 2x Applying distributive law
Cross multiplying both fractions

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

7. 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 – (Apply distributive law for 3(x-12)
2x – = 3x – (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.

8. 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 = = 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.

9. 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.