Question

solve these questions​

Answer 1

1. sum of consecutive number is 53

let the number be X,X+1

X+X+1 = 53

2X = 52

X = 26

numbers are 26 and 27

Answer 2

17.

$\blue\bigstar$ Given :

• Sum of two consecutive integers is 53.

$\pink\bigstar$ To find:

• The numbers.

$\red\bigstar$ Solution:

Let's assume the numbers be x and x + 1.

A/q,

$\sf \:x \: + \: x \: + \: 1 \: = \: 53 \\ \sf \: \implies \: 2x \: = \: 53 \: - \: 1 \\ \sf \: \implies \: x \: = \: \frac{52}{2} \\ \sf \: \implies \: \boxed{ \sf \: x = 26} \\ \therefore \: \sf \: its \: consecutive \: number \: is \: 26 \: + \: 1 \: = 27$

So, the numbers are 26 and 27.

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18.

$\blue\bigstar$ Given :

• Polynomials are m² - 2n² , 4m² - 9n² + 15 , 6m² - 5n² + 5 and m² - 4n² + 20

$\pink\bigstar$ To find:

• From the sum of m² - 2n² and 4m² - 9n² + 15 , subtract the sum of 6m² - 5n² + 5 and m² - 4n² + 20

$\red\bigstar$ Solution:

Sum of m² - 2n² and 4m² - 9n² + 15

= m² - 2n² + 4m² - 9n² + 15

= m² + 4m² - 2n² - 9n² + 15

= 5m² - 11n² + 15 ___(i)

Sum of 6m² - 5n² + 5 and m² - 4n² + 20

= 6m² - 5n² + 5 + m² - 4n² + 20

= 6m² + m² - 5n² - 4n² + 20 + 5

= 7m² - 9n² + 25 ___(ii)

Subtracting (ii) from (i),

5m² - 11n² + 15 - (7m² - 9n² + 25)

= 5m² - 11n² + 15 - 7m² + 9n² - 25

= 5m² - 7m² - 11n² + 9n² - 25 + 15

= (-2m²) - 2n² - 10