(i) x + y = 14
x – y = 4
(ii) s – t = 3
s/3 + t/2 = 6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) √2x + √3y = 0
√3x – √8y = 0
(vi) 3x/2 – 5y/3 = -2
x/3 + y/2 = 13/6
Solution:
(i) x + y = 14 and x – y = 4
From the first equation, we get,
x = 14 – y
Now, substitute the value of x in the 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) s – t = 3 and (s/3) + (t/2) = 6
From the first equation, we get,
s = 3 + t ….(1)
Now, substitute the value of s in the 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 the first 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) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3
From the first 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) √2 x + √3 y = 0 and √3 x – √8 y = 0
are the two equations.
From the first 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) (3x/2) – (5y/3) = -2 and (x/3) + (y/2) = 13/6
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
Solving this equation, we get y = 3.
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.