**Substitution Word Problems :**

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

**Example 1 : **

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

**Solution : **

Let x be the smaller number and y be the greater number.

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

Then, we have

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

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

Then, we have

4x = y + 5

Substitute (3x + 2) for y.

4x = 3x + 2 + 5

4x = 3x + 7

Subtract 3x from each side.

x = 7

Substitute 7 for x in (1).

(1)-----> y = 3(7) + 2

y = 21 + 2

y = 23

So, the numbers are 7 and 23.

**Example 2 : **

The ratio of income of two persons is 1 : 2 and the ratio of their expenditure is 3 : 7. If the first person saves $2000 and the second person saves $3000, find the expenditure of each person.

**Solution : **

From the ratio of income 1 : 2, the incomes of the two persons can be assumed as x and 2x.

From the ratio of expenditure 3 : 7, the expenditures of the two persons can be assumed as 3y and 7y.

**Savings = Income - Expenditure**

The first person saves $3000.

Then, we have

x - 3y = 3000

Add 2y to each side.

x = 3000 + 3y -----(1)

The second person saves $7000.

Then, we have

2x - 7y = 7000

Substitute (3000 + 3y) for x.

(1)-----> 2(2000 + 3y) - 7y = 3000

4000 + 6y - 7y = 3000

4000 - y = 3000

Subtract 1000 from each side.

-y = -1000

Multiply each side by -1.

y = 1000

3y = 3(1000) = 3000

7y = 7(1000) = 7000

So, the expenditures of the two persons are $3000 and $7000.

**Example 3 : **

Three chairs and two tables cost $700 and five chairs and six tables cost $1100. Find the cost of each chair.

**Solution : **

Let x and y be the costs of each chair and table respectively.

Then, we have

3x + 2y = 700 -----(1)

5x + 6y = 1700 -----(2)

Multiply (1) by 3.

(1) ⋅ 3 -----> 9x + 6y = 2100

Subtract 9x from each side.

6y = 2100 - 9x

Substitute (2100 - 9x) for 6y in (2).

(2)-----> 5x + 2100 - 9x = 1700

2100 - 4x = 1700

Subtract 2100 from each side.

-4x = -400

Divide each side ny -4.

x = 100

So, the cost of each chair is $100.

**Example 4 : **

A father's age is three times the sum of the age of his two son. After 5 years, his age will be twice the sum of the ages of his two sons. Find the present age of the father.

**Solution : **

Let x be the present age of the father and y be the sum of the present ages of the two sons.

At present :

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

After 5 years :

x + 5 = 2(y + 5 + 5)

x + 5 = 2(y + 10)

x + 5 = 2y + 20

Subtract 5 from each side.

x = 2y + 15

Substitute 3y for x.

3y = 2y + 15

Subtract 2y from each side.

y = 15

Substitute 15 for y in (1).

(1)-----> x = 3(15)

x = 45

So, the present age of the father is 45 years.

**Example 5 : **

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.

**Solution : **

Let x/y be the required fraction.

**Given : **If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1.

(x + 2) / (y + 1) = 1

Multiply each side by (y + 1).

x + 2 = 1(y + 1)

x + 2 = y + 1

Subtract 2 from each side.

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

**Given : **If the numerator is decreased by 4 and the denominator by 2, it becomes 1/2.

(x - 4) / (y - 2) = 1 / 2

Multiply each side by 2(y - 2).

2(x - 4) = 1(y - 2)

2x - 8 = y - 2

Add 8 to each side.

2x = y + 6

Substitute (y - 1) for x.

2(y - 1) = y + 6

2y - 2 = y + 6

Add 2 to each side.

2y = y + 8

Subtract y from each side.

y = 8

Substitute 8 for y in (1).

(1)-----> x = 8 - 1

x = 7

Therefore, we have

x / y = 7 / 8

So, the required fraction is 7/8.

After having gone through the stuff given above, we hope that the students would have understood, "Substitution Word Problems".

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