Solve the following pair of linear equations by the substitution method :

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.

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