**Word Problems on Linear Equations :**

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

There is a simple trick behind solving word problems using linear equations.

The picture shown below tells us the trick.

**Example 1 :**

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

**Solution :**

Let "x/y" be the fraction.

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

So, we have

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

Simplify.

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

x + 2 = y + 1

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

**Given :** In case the numerator is decreased by 4 and the denominator by 2, the fraction becomes 1/2.

So, we have

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

Simplify.

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

2x - 8 = y - 2

2x - y = 6 ------(2)

Solving (1) and (2), we get

x = 7 and y = 8

Hence, the fraction is 7/8.

**Example 2 :**

A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750 ?

**Solution :**

Let "x" be the no. of adult tickets and "y' be the no. of kids tickets sold.

**Given :**A total of 548 tickets were sold

So, we have

x + y = 548 ------(1)

**Given :** Cost of each adult ticket is $10 and kid ticket is $5 and tickets were sold for a total of $3750.

So, we have

10x + 5y = 3750

2x + y = 750 ------(2)

Solving (1) & (2), we get

x = 202 and y = 346

Hence, the number of adults tickets sold is 202 and kids tickets is 346.

**Example 3 :**

A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297.Find the number.

**Solution :**

Let "x0y" be the three digit number. (As per the given information, middle digit is zero)

Given : The sum of the other digits is 9

x + y = 9 ------(1)

By interchanging the first and third digits, the number we get is

y0x

**Given : **The number formed by interchanging the first and third digits is more than the original number by 297

y0x - x0y = 297

(100y + x) - (100x + y) = 297

100y + x - 100x - y = 297

-99x + 99y = 297

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

Solving (1) & (2), we get

x = 3 and y = 6

So, we have

x0y = 306

Hence the three digit number is 306.

**Example 4 :**

A manufacturer produces 80 units of a product at a cost of $22000 and 125 units at a cost of $28750. Assuming the cost curve to be linear, find the equation of the cost curve and then use it to estimate the cost of 95 units.

**Solution :**

Since the cost curve is linear, its equation will be

y = Ax + B.

(Here y = Total cost, x = no. of units)

**Given : **The total cost of 80 units of the product is $22000.

So, we have

22000 = 80A + B

80A + B = 22000 ------(1)

**Given : **The total cost of 125 units of the product is $28750.

So, we have

28750 = 125A + B

125A + B = 28750 ------(2)

Solving (1) and (2), we get

A = 150 and B = 10000

Then, the equation of the cost curve is

y = 150x + 10000 ------(3)

Estimate the cost of 95 units :

Plug x = 95 in (3).

(3)------> y = 150 ⋅ 95 + 10000

y = 14250 + 10000

y = 24250

Hence, the cost of 95 units is $24250.

**Example 5 :**

Y is older than X by 7 years. 15 years back X's age was 3/4 of Y's age. Find the present their present ages.

**Solution :**

Let "x" be the present age of X and "y" be the present age of Y.

**Given :** Y is older than X by 7 years.

So, we have

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

15 years back :

X's age = x - 15

Y's age = y - 15

**Given : **15 years back X's age was 3/4 of Y's age.

So, we have

(x - 15) = 3/4 ⋅ (y - 15)

Simplify.

4(x - 15) = 3(y - 15)

4x - 60 = 3y - 45

4x = 3y + 15 ------(2)

Solving (1) and (2), we get

x = 36 and y = 43

Hence, the present ages of "x" and "y" are 36 years and 43 years.

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

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