Solve the following pair of linear equations by the substitution method: (i) x+y=14; x-y=4 (ii) s-t=3; s³+t²=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) 3x²−5y³=−2; x³+y²=136

(i) x+y=14 ⇒ y=14-x
Substituting this value in the second equation, we get
x-(14-x)=4
2x=18 ⇒ x=9
Substituting this value of x in the first equation, we get
9+y=14 ⇒ y=5
(ii) s-t=3 ⇒ s=t+3
Substituting in 2nd equation
(t+3)³+t²=6
2t+6+3t⁶=6
5t+6=36 ⇒ t=6
Substituting value of t in 1st equation
s=t+3=9
(iii) ∝a₁/a₂=b₁/b₂=c₁/c₂
Hence all the points lying on the line y=3x-3 like x=2,y=3; are a solution
(iv) 0.2x+0.3y=1.3 ⇒ x=(1.3-0.3y)/0.2
Substituting this value in the second equation
0.4 × (1.3-0.3y)/0.2 + 0.5y=2.3
2.6-0.6y+0.5y=2.3
2.6-2.3=0.6y-0.5y
0.1y=0.3 ⇒ y=3
Substituting this value of y, x=(1.3-0.3y)/0.2 = (1.3-0.3 × 3)/0.2 = 0.4/0.2 =2
(v) √2x+√3y=0 ⇒ y=-√2/√3x
Substituting this in 2nd equation
√3x-√8 × (-√2/√3x)=0
√3x+4x=0 ⇒ x=0
Substituting value of x, y=-√2/√3 × 0=0
(vi) 3x²−5y³=−2; 5y³=3x²+2 ⇒ y=∛((3x²+2)/5)
Substituting this in 2nd equation
x³+(∛((3x²+2)/5))²=136
Solving this equation numerically (or by trial and error) we find x=2.
Substituting this value of x in 1st equation
y=∛((3x²+2)/5) = ∛((3(2)²+2)/5) = ∛(14/5) ≈ 1.41