Solve the following pair of linear equations by the substitution method
(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) √2 x+√3 y = 0
√3 x-√8 y = 0
(vi) (3x/2) – (5y/3) = -2
(x/3) + (y/2) = (13/6)
Answers
hope this helps u
Solution:
i) x + y = 14 ....................... (i)
x – y = 4 ….......................... (ii)
From equation (i), we get
x = 14 – y …......................... (iii)
Putting this value in equation (ii), we get
(14 – y) – y = 4
14 – 2y = 4
10 = 2y
y = 5
Putting this in equation (iii), we get
x =14 - y
x= 14 - 5
x = 9
∴ x = 9 and y = 5
(ii) s – t = 3 …........................ (i)
s/3 + t/2 = 6 …................. (ii)
From equation (i),
s= t + 3........................ ......(iii)
Putting this value in equation (ii)
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5
t= 6
Putting t=6 in equation (iii)
s= t+3
s= 6+3
s = 9
∴ s = 9, t = 6
(iii) 3x – y = 3 …............................ (i)
9x – 3y = 9 …......................... (ii)
From equation (i)
y = 3x – 3 …........................... (iii)
Putting this value in equation (ii)
9x – 3(3x – 3) = 9
9x – 9x + 9 = 9
9 = 9
This statement has no variable but I it is a true statement for all values of x,
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x – 3
Therefore, one of its possible solutions is x = 1, y =0
(iv) 0.2x + 0.3y = 1.3 …................. (i)
0.4x + 0.5y = 2.3 …................. (ii)
On multiplying both equation by 10 we get
2x + 3y = 13........................(iii)
4x+5y=23...........................(iv)
From eq iii
3y = 13- 2x
y=( 13-2x) /3................(v)
On substituting the value of y in equation iv
4x + 5/3(13 -2x) =23
(4x×3 + 5(13-2x))/3=23
12x+ 65 -10x = 23×3
12x-10x= 69-65
2x= 4
x= 4/2= 2
x= 2
On substituting x = 2 in eq v,
y= (13- 2×2)/3=( 13-4)/3= 9/3= 3
y=3
∴ x = 2 and y = 3
