**Solving Word Problems Involving Two Numbers :**

In the section, we will learn how to solve word problems involving two numbers.

**Example 1 :**

The sum of the number and its positive square root is 7/36. Find the number.

**Solution :**

Let x be the required number

The sum of the number and its positive square root is 7/36.

x + √x = 7/36

√x = (7/36) - x

By taking squares on both sides, we get

x = [(7/36) - x]^{2}

x = (7/36)^{2} - 2 (7/36) x + x^{2}

x = (49/1296) - (7x/18) + x^{2}

1296x = 49 - 504 x + 1296x^{2}

1296x^{2} - 504x - 1296x + 49 = 0

1296x^{2} - 1800x + 49 = 0

(36x - 1) (36x - 49) = 0

36x - 1 = 0 x = 1/36 |
36x - 49 = 0 36x = 49 x = 49/36 |

Therefore the required number is 1/36

**Verification :**

The sum of the number and its positive square root is 7/36.

(1/36) + √(1/36) = 7/36

(1/36) + (1/6) = 7/36

taking L.C.M we get

(1 + 6)/36 = 7/36

7/36 = 7/36

**Example 2 :**

The sum of two numbers is 15 and the sum of its reciprocals is 3/10. Find the numbers.

**Solution :**

Let x and y are the required two numbers

The sum of two numbers is 15

x + y = 15

y = 15 - x -----(1)

the sum of its reciprocals is 3/10

(1/x) + (1/y) = 3/10

(y + x)/xy = 3/10

10 (x + y) = 3xy

10x + 10y = 3xy ---- (2)

Now we are going to apply the value of y in the second equation

10x + 10(15 - x) = 3x(15 - x)

10x + 150 - 10x = 45x - 3x^{2}

3x^{2} - 45x + 150 = 0

By dividing the entire equation by 3, we get

x^{2} - 15x + 50 = 0

(x - 10) (x - 5) = 0

x - 10 = 0 x = 10 |
x - 5 = 0 x = 5 |

Therefore the two numbers are 10 and 5.

**Verification :**

The sum of two numbers is 15

10 + 5 = 15

15 = 15

the sum of its reciprocals is 3/10

(1/10) + (1/5) = 3/10

Taking l.C.M we get,

(1 + 2)/10 = 3/10

3/10 = 3/10

After having gone through the stuff given above, we hope that the students would have understood how to solve word problems with two numbers.

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