Question : Solve the following pair of linear equations by the substitution method :
(i) x + y = 14 and x – y = 4
(ii) s – t = 3 and \(s\over 3\) + \(t\over 2\) = 6
(iii) 3x – y = 3 and 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3
(v) \(\sqrt{2}x + \sqrt{3}y\) = 0 and \(\sqrt{3}x – \sqrt{8}y\) = 0
(vi) \(3x\over 2\) – \(5y\over 3\) = 2 and \(x\over 3\) + \(y\over 2\) = \(13\over 6\)
Solution :
(i) The given system of equations is
x + y = 14 …….(1)
and x – y = 4 ……(2)
From equation (1), y = 14 – x
Substituting y = 14 – x in equation (2), we get
x – (14 – x) = 14 \(\implies\) x – 14 + x = 4
\(\implies\) 2x = 4 + 14 \(\implies\) 2x = 18
\(\implies\) x = 9
Putting x = 9 in equation (1), we get
9 + y = 14 \(\implies\) y = 5
Hence, the solution of the given system of linear equations is x = 9 and y = 5.
(ii) The given system of equations is
s – t = 3 ……(1)
and \(s\over 3\) + \(t\over 2\) = 6 …..(2)
From equation (1), s = 3 + t
Substituting s = 3 + t in equation (2), we get
\(3 + t\over 3\) + \(t\over 2\) = 6 \(\implies\) 2(3 + t) + 3t = 36
\(\implies\) 6 + 2t + 3t = 36 \(\implies\) 5t = 30 \(\implies\) t = 6
Putting t = 6 in equation (1), we get
s – 6 = 3 \(\implies\) s = 9
Hence, the solution of the given system of linear equations is s = 3 and t = 6.
(iii) The given system of equations is
3x – y = 3 …….(1)
and 9x – 3y = 9 ……(2)
From equation (1), y = 3x – 3
Substituting y = 3x – 3 in equation (2), we get
9x – 3(3x – 3) = 9 \(\implies\) 9x – 9x + 9 = 9
or 9 = 9
Since, this statement always remains true for all value of x.
Therefore, Equation (1) and (2) have infinitely many solutions.
(iv) The given system of equations is
0.2x + 0.3y = 1.3 \(\implies\) 2x + 3y = 13 …….(1)
and 0.4x + 0.5y = 2.3 \(\implies\) 4x + 5y = 23 ……(2)
From equation (2), 5y = 23 – 4x \(\implies\) y = \(23 – 4x\over 5\)
Substituting y = \(23 – 4x\over 5\) in equation (1), we get
10x + 69 – 12x = 65 \(\implies\) -2x = -4
\(\implies\) x = 2
Putting x = 2 in equation (1), we get
3y = 13 – 4 \(\implies\) y = 3
Hence, the solution of the given system of linear equations is x = 2 and y = 3.
(v) The given system of equations is
\(\sqrt{2}x + \sqrt{3}y\) = 0 …….(1)
and \(\sqrt{3}x – \sqrt{8}y\) = 0 ……(2)
From equation (2), y = \(\sqrt{3}x\over \sqrt{8}\)
Substituting y = \(\sqrt{3}x\over \sqrt{8}\) in equation (1), we get
\(\sqrt{2}x + \sqrt{3y}({\sqrt{3}x\over \sqrt{8}})\) = 0 \(\implies\) \(\sqrt{2}x\) + \(3x\over \sqrt{8}\) = 0
\(\implies\) 4x + 3x = 0 \(\implies\) 7x = 0
\(\implies\) x = 0
Putting x = 0 in equation (1), we get
0 + \(\sqrt{3}y\) = 14 \(\implies\) y = 0
Hence, the solution of the given system of linear equations is x = 0 and y = 0.
(vi) The given system of equations is
\(3x\over 2\) – \(5y\over 3\) = 2 \(\implies\) 9x – 10y = -12 ……(1)
and \(x\over 3\) + \(y\over 2\) = \(13\over 6\) \(\implies\) 2x + 3y = 13 ……(2)
From equation (1), 10y = 9x + 12 \(\implies\) y =\(9x + 12\over 10\)
Substituting y =\(9x + 12\over 10\) in equation (2), we get
20x + 27x + 36 = 130 \(\implies\) 47x = 130 – 36
\(\implies\) 47x = 94 \(\implies\) x = 2
Putting x = 2 in equation (1), we get
-10y = -12 – 18 \(\implies\) -10y = -30
Hence, the solution of the given system of linear equations is x = 2 and y = 3.
