Question

घटनचा
(ii) सोडवा : 2x-3y = 9; 2x+y= 13.
(iv) 10 ते 250 मधील 4 ने भाग जाणाऱ्या नैसर्गिक संख्या किती?
या वर्गसमीकरणाची मुळे वास्तव व समान असल्यास kच
40​

Answer 1

Question 1) :- Solve the event :- 2x-3y = 9; 2x + y = 13.

Answer :-

→ 2x - 3y = 9 ------- Eqn.(1)

→ 2x + y = 13 ------- Eqn.(2)

subtracting Eqn.(1) from Eqn.(2),

→ (2x + y) - (2x - 3y) = 13 - 9

→ 2x - 2x + y + 3y = 4

→ 4y = 4

→ y = 1 .

putting value of y in Eqn.(2),

→ 2x + 1 = 13

→ 2x = 13 + 1

→ 2x = 14

→ x = 7 .

Question 2) :- What is the sum natural number divisible by 4 from 10 to 250 ?

Solution :-

• first term divisible by 4 = 12 = a
• last term = 248 = an
• common difference = 4 .

so,

→ an = a + (n - 1)d

→ 248 = 12 + (n - 1)4

→ 248 = 12 + 4n - 4

→ 248 = 8 + 4n

→ 248 - 8 = 4n

→ 240 = 4n

→ n = 60 .

then,

→ sn = (n/2)[first term + last term]

→ sn = (60/2)[12 + 248]

→ sn = 30 * 260

→ sn = 7800 (Ans.)

Learn more :-

Netra a std X student is fond of reading

books. She wants to buy a book that costs

₹207.50. She has decided to save ...

https://brainly.in/question/28900514