Word Problems on Linear Equations

Ever wondered how to solve everyday math puzzles like "How old will I be?" or "How much money do I need?" That's exactly what linear equations help us figure out!

Think of a linear equation like a balance scale:

• One side has some known numbers and an unknown (let's call it x)

• The other side has a known number

• Our job is to find what x equals to keep both sides balanced!

Example: If x + 5 = 12, then x must be 7 (because 7 + 5 = 12)

Problem: There are some pedhas in a box. If some children are given 2 pedhas each, the pedhas would be enough for 20 children. How many pedhas are there in the box?

Solution:

Step 1: Define Variable
Let the total number of pedhas be p.

Step 2: Form Equation
`p/2` = 20

Step 4: Solve and Conclude
∴ `p/2` × 2 = 20 × 2    (Multiplying both sides by 2)
p = 40
Therefore, there are 40 pedhas in the box.

Problem: One-third of a number added to one-fifth of it gives 32. Find the number.

Solution:

Step 1: Define Variable
Let the required number be x.

Step 2: Form Equation
`1/3`x + `1/5`x = 32

Step 3: Solve (LCM is 15)
⇒          `"5x + 3x"/15` = 32
⇒          `"8x"/15` = 32 and x = 32 

Step 4: Isolate x
x = 32 × `15/8` = 60     
∴ The required number is 60.

Problem: A number is decreased by 15, and the new number so obtained is multiplied by 3; the result is 81. Find the number.

Solution:
Step 1: Define Variable
Let the number be x.

Step 2: Form Equation
The number decreased by 15 = x - 15
The new number (x - 15)
multiplied by 3 = 3(x − 15)

Step 3: Solve
Given: 3(x − 15) = 81 => 3x − 45 = 81
=> 3x = 81 + 45 = 126 

Step 4: Isolate x
=> x = `126/3` = 42
∴ The required number is 42.

• Equations are Balances

• Transposition is Efficient: Use transposition (changing the sign when moving a term) to quickly isolate the variable.

• Structure is Key: Always follow the
  4-Step Master Plan (Read → Variable → Equation → Solve)