**Word Problems on Simultaneous Linear Equations :**

Solving word problems using simultaneous 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 :**

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

**Solution :**

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

x > y

**Given : **One number is greater than thrice the other number by 2.

So, we have

x = 3y + 2 ------(1)

**Given :** 4 times the smaller number exceeds the greater by 5

So, we have

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

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

(2)------> 4y - (3y + 2) = 5

Simplify.

4y - 3y - 2 = 5

y - 2 = 5

Add 2 to both sides.

y = 7

Plug y = 7 in (1).

x = 3 ⋅ 7 + 2

x = 21 + 2

x = 23

Hence the two numbers are 23 and 7.

**Example 2 :**

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

**Solution :**

Let "xy" be the two digit number.

**Given : ** The two digit number is seven times the sum of its digits.

xy = 7(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 = 7(x + y)

10x + y = 7x + 7y

Simplify.

3x - 6y = 0

Divide both sides by 3.

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

**Given : ** The number formed by reversing the digits is 18 less than the given number.

The number formed by reversing the digits is

yx

yx is 18 less than xy.

So, we have

xy - yx = 18

[10 ⋅ x + 1 ⋅ y] - [10 ⋅ y + 1 ⋅ x] = 18

(10x + y) - (10y + x) = 18

Simplify.

10x + y - 10y - x = 18

9x - 9y = 18

Divide both sides by 9.

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

Solving (2) and (3), we get

x = 4 and y = 2

Hence, the two digit number is 42.

**Example 3 :**

In a rectangle, if the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm . If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm . Find the area of the rectangle.

**Solution :**

Let "x" and "y" be the length and width of the rectangle respectively.

Then, area of the rectangle is

= x ⋅ y

**Given :** If the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm

So, we have

(x + 2) ⋅ (y - 2) = x ⋅ y - 28

Simplify.

xy - 2x + 2y - 4 = xy - 28

- 2x + 2y = - 24

Divide both sides by 2.

- x + y = - 12 ------(1)

**Given : **If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm

So, we have

(x - 1) ⋅ (y + 2) = x ⋅ y + 33

Simplify.

xy + 2x - y - 2 = xy + 33

2x - y = 35 ------(2)

Solving (1) and (2), we get

x = 23 and y = 11

Area of the original rectangle is

= x ⋅ y

= 23 ⋅ 11

= 253 sq.cm

Hence the area of the rectangle is 253 sq.cm

**Example 4 :**

Sum of incomes of A and B is $2640. If B's income is 20% more than A, find the income of A and B.

**Solution :**

Let "x" and "y" be the incomes of A and B respectively.

Then, we have

x + y = 2640 ------(1)

**Given : **B's income is 20% more than A.

So, we have

y = 120% of x

y = 1.2x ------(2)

Plug y = 1.2x in (1).

x + 1.2x = 2640

Simplify.

2.2x = 2640

Divide both sides by 2.2.

x = 1200

Plug x = 1200 in (1)

y = 1.2 ⋅ 1200

y = 1440

Hence the income of A is $1200 and B is $1440.

**Example 5 :**

Sum of the cost price of two products is $50. Sum of the selling price of the same two products is $52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.

**Solution :**

Let "x" and "y" be the cost prices of two products.

Then, we have

x + y = 50 ------(1)

Let us assume that "x" is sold at 20% profit.

Then, the selling price of "x" is

= 120% of "x"

= 1.2x

Let us assume thatr "y" is sold at 20% loss.

Then, the selling price of "y" is

= 80% of "y"

= 0.8y

**Given :** Sum of the selling price of the same two products is $52.

So, we have

1.2x + 0.8y = 52

Multiply both sides by 10.

12x + 8y = 520

Divide both sides by 4.

3x + 2y = 130 --------(2)

Solving (1) and (2), we get

x = 30 and y = 20

Hence the cost prices of two products are $30 and $20.

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