Question

Solve the following pair of linear equations by the substitution method.
$
  {\text{(i)}}{\text{. }}x + y = 14{\text{;}}x - y = 4 \\
  {\text{(ii)}}{\text{. }}s - t = 3{\text{;}}\dfrac{s}{3}{\text{ + }}\dfrac{t}{2}{\text{ = 6 }} \\
  {\text{(iii)}}{\text{. }}3x - y = 3{\text{ ; 9}}x - 3y = 9 \\
  {\text{(iv)}}{\text{. 0}}{\text{.2}}x + 0.3y = 1.3{\text{ ; 0}}{\text{.4}}x + 0.5y = 2.3{\text{ }} \\
  {\text{(v)}}{\text{. }}\sqrt 2 x + \sqrt 3 {\text{y = 0; }}\sqrt 3 x - \sqrt 8 y = 0 \\
  {\text{(vi)}}{\text{. }}\dfrac{{3x}}{2} - \dfrac{{5y}}{3}{\text{ = }} - {\text{2 ;}}\dfrac{x}{3}{\text{ + }}\dfrac{y}{2}{\text{ = }}\dfrac{{13}}{6}{\text{ }} \\
 $

Answer 1

Hint: In this question convert given equations into some simple forms if required then use the substitution method to find the variable. Substitution Method: It is the algebraic method to solve simultaneous linear equations. In this method, the value of one variable from one equation is substituted in the other equation. In this way, a pair of the linear equation gets transformed into one linear equation with only one variable which can then easily be solved.

Complete step-by-step solution -
${\text{(i)}}{\text{. }}x + y = 14{\text{;}}x - y = 4$
Let

  $ \Rightarrow {\text{ }}x + y = 14 $ …………………..eq.(1)
  $ \Rightarrow x - y = 4 $ ……………………………. eq.(2)

From eq.2, on rearranging it, we get
 $ \Rightarrow x = 4 + y $ …………………. eq.(3)

Put value of x from eq.3 in eq.1, we get
$
   \Rightarrow (y + 4) + y = 14 \\
   \Rightarrow 2y{\text{ }} = {\text{ 10}} \\
   \Rightarrow y{\text{ }} = {\text{ 5}} \\
$
Now put value of y in eq.3 we get
$ \Rightarrow x = 9 $
Hence, $ x = 9,{\text{ y = 5}} $

$ {\text{(ii)}}{\text{. }}s - t = 3{\text{;}}\dfrac{s}{3}{\text{ + }}\dfrac{t}{2}{\text{ = 6 }} $
$ \Rightarrow s - t = 3 $ ………………….. eq.(1)
$ \Rightarrow {\text{ }}\dfrac{s}{3}{\text{ + }}\dfrac{t}{2}{\text{ = 6 }} \\ $

Now, we take LCM in LHS, we get
$ \Rightarrow {2s + 3t = 36} $ ……………………… eq.(2)
From eq.1, on rearranging its terms, we get
 $ \Rightarrow s = 3 + t $ ………………………… eq.(3)

Put value of s from eq.3 in eq.2, we get
$
   \Rightarrow {\text{ 2(3 + t) + 3t = 36}} \\
   \Rightarrow 5t = 30 \\
   \Rightarrow t = 6 \\
 $
Now put value t in eq.3 we get
$ \Rightarrow s = 9 $
Hence, $ s = 9,{\text{ t = 6}}$

${\text{(iii)}}{\text{. }}3x - y = 3{\text{ ; 9}}x - 3y = 9$
$ \Rightarrow {\text{ 3}}x - y = 3 $ ……………. eq.(1)
$ \Rightarrow 9x - 3y = 9 $ ………………. eq.(2)
From eq.2
$ \Rightarrow y = 3x - 3 $ …………………. eq.(3)
Put value of y from eq.3 in eq.1, we get
$
   \Rightarrow 3x - 3x + 3 = 3 \\
   \Rightarrow 3 = 3 \\
$
Hence, equations have infinite solutions

${\text{(iv)}}{\text{. 0}}{\text{.2}}x + 0.3y = 1.3{\text{ ; 0}}{\text{.4}}x + 0.5y = 2.3$
$ \Rightarrow {\text{ 0}}{\text{.2}}x + 0.3y = 1.3$
Multiply above eq. with 10, we get
$2x + 3y = 13 $ ……………….. eq.(1)
$ \Rightarrow 0.4x - 0.5y = 2.3 $
Multiply above eq. with 10, we get
$4x + 5y = 23 $ ………………….. eq.(2)
From eq.1
 $ \Rightarrow x = \dfrac{{13 - 3y}}{2} $ ………………. eq.(3)
Put value of x from eq.3 in eq.2, we get
 $ \Rightarrow 4(\dfrac{{13 - 3y}}{2}{\text{)}} + 5y = 23{\text{ }} \\ $
  $ \Rightarrow y{\text{ }} = {\text{ 3}} \\ $
Now put value y in eq.3 we get
$ \Rightarrow x = 2$
Hence, $x = 2,{\text{ y = 3}}$

${\text{(v)}}{\text{. }}\sqrt 2 x + \sqrt 3 {\text{y = 0; }}\sqrt 3 x - \sqrt 8 y = 0$
$ \Rightarrow {\text{ }}\sqrt 2 x + \sqrt 3 y = 0 $ ……………… eq.(1)
$ \Rightarrow \sqrt 3 x - \sqrt 8 y = 0 $ ………….. eq.(2)
 From eq.1
 $ \Rightarrow x = - \sqrt {\dfrac{3}{2}} y $ ……………… eq.(3)
Put value of x from eq.3 in eq.2, we get
$
   \Rightarrow \sqrt 3 ( - \sqrt {\dfrac{3}{2}} {\text{y}}) - \sqrt 8 y = 0 \\
   \Rightarrow y{\text{ }} = {\text{ 0}} \\
$
Now put value y in eq.3 we get
$ \Rightarrow x = 0$
Hence, $x = 0,{\text{ y = 0}}$

${\text{(vi)}}{\text{. }}\dfrac{{3x}}{2} - \dfrac{{5y}}{3}{\text{ = }} - {\text{2 ;}}\dfrac{x}{3}{\text{ + }}\dfrac{y}{2}{\text{ = }}\dfrac{{13}}{6}$
$ \Rightarrow \dfrac{{3x}}{2} - \dfrac{{5y}}{3}{\text{ = }} - {\text{2}}$
Multiply above eq. with 6, we get
$9x - 10y = - 12 $ ……………….. eq.(1)
$\dfrac{x}{3}{\text{ + }}\dfrac{y}{2}{\text{ = }}\dfrac{{13}}{6}$
Multiply above eq. with 6, we get
${\text{2}}x + 3y = 13 $ ……………….. eq.(2)

From eq.2
 $ \Rightarrow x = \dfrac{{13 - 3y}}{2} $ …………… eq.(3)
Put value of x from eq.3 in eq.1, we get

$
   \Rightarrow 9(\dfrac{{13 - 3y}}{2}{\text{)}} - {\text{10}}y = - 12{\text{ }} \\
   \Rightarrow y = 3 \\ $
Now put value y in eq.3 we get
$ \Rightarrow x = 2$
Hence, $x = 2,{\text{ y = 3}}$

Note: Whenever you get this type of question the key concept to solve is to make the given equation into some simpler forms. Then find the value of one of the unknown variables in terms of another variable. Then put this value in one of given equations to get the value of another variable. And similarly find the other unknown variable.