Question

Solve the following pair of linear equations by the substitution method. 
(i) x + y = 14 
    x – y = 4   

(ii) s – t = 3 
    \(\frac{s}{3} + \frac{t}{2}\) =6 

(iii) 3x – y = 3 
      9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3 
     0.4x + 0.5y = 2.3 

(v)\(\sqrt2x\) + \(\sqrt3y\)=0
    \(\sqrt3x\) - \(\sqrt8y\) = 0

(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
    \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

Answer 1

(i) x + y = 14 ……………………..(1) 
x − y = 4        ..…………………...(2)

From (1), we obtain 
x = 14 − y       .……………..........(3) 

Substituting this value in equation (2), we obtain 
(14-y) -y =4
14 -2y = 4
10=2y
y=5

Substituting this in equation (3), we obtain 
x=9 
∴ x=9, y=5

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(ii) s-t =3      ..……………..(1) 
     \(\frac{s}{3} + \frac{t}{2}\) =6 ……………….(2)

From (1), we obtain 
s= t+3 ………..(3) 

Substituting this value in equation (2), we obtain
\(\frac{t+3}{3}\) + \(\frac{t}{2}\) =6
2t +6 +3t = 36 
5t =30
  t = 6              
Substituting in equation (3), we obtain
s=9
∴s=9 , t=6

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(iii) 3x − y = 3 ......................(1)
      9x − 3y = 9 .....................(2)

From (1), we obtain 
y = 3x − 3 ......................(3)

Substituting this value in equation (2), we obtain 
9x -3 (3x-3) =9 
9x -9x +9 =9
9 = 9 

This is always true. 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.

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(iv) 0.2x + 0.3y = 1.3 ........................(1)
      0.4x + 0.5y = 2.3 ........................(2)

From equation (1), we obtain 
x=\(\frac{1.3-0.3y}{0.2}\) ........................(3)

Substituting this value in equation (2), we obtain 
0.4(\(\frac{1.3-0.3y}{0.2}\)) + 0.5 y = 2.3

2.6 -0.6y + 0.5 y =2.3
2.6 -2.3= 0.1y
03=0.1y
y=3

Substituting this value in equation (3), we obtain 
x= \(\frac{1.3 -0.3 \times 3 }{ 0.2 }\)

  = \(\frac{1.3 -0.9}{0.2}\) =\(\frac{ 0.4}{0.2}\) =2

∴ x=2, y=3

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(v) \(\sqrt2x + \sqrt3y\) =0  .................(1)
      \(\sqrt3x - \sqrt8y\) =0 .................(2)

from equation (1), we obtain
x= -\(\frac{\sqrt3y}{\sqrt2}\)                    .....................(3)

Substituting this value in equation (2), we obtain
\(\sqrt3(\frac{-\sqrt3y}{\sqrt2}) -\sqrt8y\)=0 

\(\frac{-3y}{\sqrt2} -2\sqrt2y\) =0

y(\(\frac{-3y}{\sqrt2} -2\sqrt2y\)) =0

y=0

Substituting this value in equation (3), we obtain
x = 0 
 ∴ x = 0, y = 0

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(vi) \(\frac{3}{2}x - \frac{5}{3}y\) = -2 ...............(1)
      \(\frac{x}{3} + \frac{y}{2 }\)=\(\frac{13}{6}\)       ................(2)

From equation (1), we obtain 
9x-10y =12
x=\(\frac{-12+10y}{9}\)               ...............(3)

Substituting this value in equation (2), we obtain 
\(\frac{-12+\frac{10y}{9}}{3} +\frac{y}{2}\) =\(\frac{13}{6}\)

-\(\frac{12+10y}{27 }\)+\(\frac{y}{2}\) = \(\frac{13}{6}\)

\(\frac{-24 +20y +27y}{54}\) = \(\frac{13}{6}\)

\(47y\)= \(117 + 24\)
\(47y\) =\(141\)
    \(y\) = \(3\) 

Substituting this value in equation (3), we obtain 
x= \(\frac{-12 +10 \times 3}{9 }\)= \(\frac{18}{9}\) =2

∴ x=2, y=3