**Word Problems on Simultaneous Linear Equations :**

In this section, we will learn, how to solve word problems using simultaneous equations.

**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.

Then, we have

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

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

Then, we have

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

From (1), substitute (3y + 2) in (2).

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

Simplify.

4y - 3y - 2 = 5

y - 2 = 5

Add 2 to each side.

y = 7

Substitute 7 for y in (1).

x = 3(7) + 2

x = 21 + 2

x = 23

So, 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)

In the two digit number xy, x is in tens place and y is in ones place.

Then, we have

xy = 10 ⋅ x + 1 ⋅ y

xy = 10x + y

Substitute (10x + y) for xy in (1).

(1)-----> 10x + y = 7(x + y)

10x + y = 7x + 7y

Simplify.

3x - 6y = 0

Divide each side 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.

Then, 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 each side by 9.

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

Solving (2) and (3), we get

x = 4 and y = 2

So, the required 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

Then, we have

(x + 2) ⋅ (y - 2) = x ⋅ y - 28

Simplify.

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

- 2x + 2y = - 24

Divide each side 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

Then, 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

So, 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.

Then, we have

y = 120% of x

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

Substitute 1.2x for y in (1).

x + 1.2x = 2640

Simplify.

2.2x = 2640

Divide each side by 2.2.

x = 1200

Substitute 1200 for x in (2)

y = 1.2 ⋅ 1200

y = 1440

So, 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.

**Given :** Sum of the cost prices of the two products is $50.

Then, we have

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

**Given :** One is sold at 20% profit and other one is sold at 20% loss.

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 that 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 two products is $52.

Then, we have

1.2x + 0.8y = 52

Multiply both sides by 10.

12x + 8y = 520

Divide each side by 4.

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

Solving (1) and (2), we get

x = 30 and y = 20

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

After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems using simultaneous linear equations.

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