Pawan Prajapati
Last Activity: 3 Years ago
Solutions:
(i) 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
Now, substitute the value of t in equation (1)
s = 3 + 6 = 9
Therefore, s = 9 and t = 6.
(iii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3+y)/3
Now, substitute the value of x in the given second equation to get,
9(3+y)/3 – 3y = 9
⇒9 +3y -3y = 9
⇒ 9 = 9
Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.
(iv) Given,
0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.
From 1st equation, we get,
x = (1.3- 0.3y)/0.2 _________________(1)
Now, substitute the value of x in the given second equation to get,
0.4(1.3-0.3y)/0.2 + 0.5y = 2.3
⇒2(1.3 – 0.3y) + 0.5y = 2.3
⇒ 2.6 – 0.6y + 0.5y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3
Now, substitute the value of y in equation (1), we get,
x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2
Therefore, x = 2 and y = 3.
(v) Given,
√2 x + √3 y = 0 and √3 x – √8 y = 0
are the two equations.
From 1st equation, we get,
x = – (√3/√2)y __________________(1)
Putting the value of x in the given second equation to get,
√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0
⇒ y = 0
Now, substitute the value of y in equation (1), we get,
x = 0
Therefore, x = 0 and y = 0.
(vi) Given,
(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.
From 1st equation, we get,
(3/2)x = -2 + (5y/3)
⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)
Putting the value of x in the given second equation to get,
((-12+10y)/9)/3 + y/2 = 13/6
⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6 Now, substitute the value of y in equation (1), we get,
(3x/2) – 5(3)/3 = -2
⇒ (3x/2) – 5 = -2
⇒ x = 2
Therefore, x = 2 and y = 3.
