**Word Problems Involving Linear Equations :**

Solving word problems on linear equations is sometimes a difficult job for some students.

Actually it is not. There is a simple trick behind it.

The picture given below tells us the trick.

**Example 1 :**

Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers.

**Solution :**

Let "x" and "y" be the two numbers such that

x > y

**Given :** The sum of the two number is 16.

So, we have

x + y = 16 ------(1)

**Given :** 1/5th of a the greater equal to 1/3rd of the smaller

So, we have

1/5 ⋅ x = 1/3 ⋅ y

Simplify.

3x = 5y ------(2)

Solving (1) and (2), we get

x = 10 and y = 6

Hence, the two numbers are 10 and 6.

**Example 2 :**

The wages of 8 men and 6 boys amount to $33. If 4 men earn $4.50 more than 5 boys, determine the wages of each man and boy.

**Solution :**

Let "x" and "y" be the wages of each man and boy.

**Given :** The wages of 8 men and 6 boys amount to $33.

So, we have

8x + 6y = 33 ------(1)

**Given :** 4 men earn $4.50 more than 5 boys.

So, we have

4x - 5y = 4.50 ------(2)

Solving (1) and (2), we have

x = 3 and y = 1.5

Hence, the wages of each man and each boy are $3 and $1.50 respectively.

**Example 3 :**

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number.

**Solution :**

Because the number is between 10 and 100, it has to be a two digit number.

So, let "xy" be the number between 10 and 100.

**Given : ** The number between 10 and 100 is five times the sum of its digits.

xy = 5(x + y) ------(1)

xy is a two digit number.

And x is in tens place and y is in ones place.

So, we have

xy = 10 ⋅ x + 1 ⋅ y

xy = 10x + y

Then, we have

(1)------> 10x + y = 5(x + y)

10x + y = 5x + 5y

Simplify.

5x - 4y = 0 -------(2)

**Given : ** If 9 be added to it the digits are reversed.

xy + 9 = yx

10 ⋅ x + 1 ⋅ y + 9 = 10 ⋅ y + 1 ⋅ x

10x + y + 9 = 10y + x

Simplify.

9x - 9y = - 9

Divide both sides by 9.

x - y = - 1 ------(3)

Solving (2) and (3), we get

x = 4 and y = 5

Hence, the required number is 45.

**Example 4 :**

The age of a man is three times the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man.

**Solution :**

Let "x" be the present age of the man and "y" be the sum of the present ages of his two sons. **Given :** Present age of the man is 3 times the sum of the ages of 2 sons.

So, we have

x = 3y ------(1)

5 years hence,

Age of the man = x + 5

Sum of the ages of his two sons = y + 5 + 5 = y + 10

(There are two sons in y. So 5 is added twice)

**Given :** 5 years hence, age of the man will be double the sum of the ages of his two sons.

So, we have

x + 5 = 2(y + 10)

Simplify.

x + 5 = 2y + 20

x = 2y + 15 ------(2)

From (1), we can plug x = 3y in (2).

(2)------> 3y = 2y + 15

Subtract 2y from both sides.

y = 15

Plug y = 15 in (1).

(1)------> x = 3 ⋅ 15

x = 45

Hence the present age of the man is 45 years.

**Example 5 :**

A trader has 100 units of a product. He sells some of the units at $6 per unit and the remaining units at $8 per units. He receives a total of $660 for all 100 units. Find the number units sold in each category.

**Solution :**

Let x be the no. of units sold at $6/unit and y be the no. of units sold at $8/unit.

**Given :** The trader sells 100 units in all.

So, we have

x + y = 100 ------(1)

**Given :** He receives a total of $660 for all 100 units.

So, we have

6x + 8y = 660

Divide both sides by 2.

3x + 4y = 330 ------(2)

Solving (1) and (2), we get

x = 70 and y = 30

Hence, the no. of tickets sold at $6 per unit is 70 and the no. of tickets sold at $8 per unit is 30.

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