Ques No. 3 and 4 Exercise 3.3 Maths Class 10th
Answers
Answered by
5
Answer:
Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get,
x = 14 – y
Now, substitute the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the exact value of x;
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.
(ii) Given,
s – t = 3 and (s/3) + (t/2) = 6 are the two equations.
From 1st equation, we get,
s = 3 + t ________________(1)
Now, substitute the value of s in second equation to get,
(3+t)/3 + (t/2) = 6
⇒ (2(3+t) + 3t )/6 = 6
⇒ (6+2t+3t)/6 = 6
⇒ (6+5t) = 36
⇒5t = 30
⇒t = 6
Similar questions