# Linear Equations in Two Variables : Exercise 4.2 (Mathematics NCERT Class 9th) Q.1      Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution
(ii) only two solution
(iii) infinitely many solutions
Sol.

Given equation is y = 3 x + 5
For y = 0 , 3x + 5 = 0 $\Rightarrow \,x = - {5 \over 3}$
Therefore $\left( { - {5 \over 3},0} \right)$ is one of solution.
For x = 0, y = 0 + 5 = 5
Therefore, (0, 5) is another solution.
For x = 1, y = 3 × 1 + 5 = 8
Therefore (1, 8) is another solution.
Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with
this value of y constitutes another solution of the given equation. So, there is no end to different
solutions of a linear equation in two variables.
Therefore, A linear equation in two variables has infinitely many solutions.

Q.2      Write four solutions for each of the following equations :
(i) 2x + y = 7
(ii) $\pi x + y = 9$
(iii) x = 4y

Sol.

(i) The given equation can be written as y = 7 – 2x.
For x = 0 , y = 7 – 2 × 0 = 7 – 0 = 7
For x = 1, y = 7 – 2 × 1 = 7 – 2 = 5
For x = 2, y = 7 – 2 × 2 = 7 – 4 = 3
For x = 3, y = 7 – 2 × 3 = 7 – 6 = 1
Therefore the four solutions of the given equation are (0, 7), (1, 5) , (2, 3) and (3, 1) .
(ii) The given equation can be written as  $y = 9 - \pi x$

For x = 0, y = 9 – 0 = 9
For x = 1, $y = 9 - \pi$
For x = 2 , $y = 9 - 2\pi$
For x = 3, $y = 9 - 3\pi$
Therefore the four solutions of the given equation are (0, 9),
(1,9$- \pi$) , (2,9 $- 2\pi$) and (3, 9 $- 3\pi$)
(iii) The given equation can be written as x = 4y

For x = 0,   y = 0
For  x = 1 ;  $y = {1 \over 4}$
For x = 2 ;  $y = {2 \over 4} = {1 \over 2}$
For x = 3 ; $y = {3 \over 4} = {3 \over 4}$
Therefore the four solutions of the given equation are $\left( {0,0} \right)\,\left( {1,{1 \over 4}} \right)\left( {2,{1 \over 2}} \right)\,and\,\left( {3,{3 \over 4}} \right)$

Q.3      Check which of the following are solutions of the equation x – 2y = 4 and which are not :
(i) (0, 2)     (ii) (2, 0)      (iii) (4, 0)       (iv) $\left( {\sqrt 2 ,\,4\sqrt 2 } \right)$(v) (1, 1)
Sol.

(i) Putting x = 0 , y = 2 in L.H.S. of x – 2y = 4, we have
L.H.S. = 0 – 2 × 2 = – 4 $\ne$ R.H.S.
Therefore x = 0 , y = 2 is not its solution.
(ii) Putting x = 2, y = 0 in L.H.S. of x – 2y = 4, we have

L.H.S. = 2 – 2 × 0 = 2 – 0 = 2 $\ne$ R.H.S.
Therefore x = 2, y = 0 is not its solution.
(iii) Putting x = 4, y = 0 in the L.H.S. of x – 2y = 4, we have

L.H.S. = 4 – 0 = 4 = R.H.S.
Therefore x = 4, y = 0 is its solution.
(iv) Putting $x = \sqrt 2 ,\,y = 4\sqrt 2$ in the L.H.S. of x – 2y = 4, we have

L.H.S. $x = \sqrt 2 - 2 \times 4\sqrt 2 = \sqrt 2 - 8\sqrt 2$
$= - 7\sqrt 2 \, \ne$ R.H.S.
Therefore $\left( {\sqrt 2 ,4\sqrt 2 } \right)$ is not its solution.
(v) Putting x = 1, y = 1 in the L.H.S. of x – 2y = 4, we have

L.H.S. = 1 – 2 × 1 = 1 – 2 = – 1 $\ne$ R.H.S.
Therefore x = 1, y = 1 is not its solution.

Q.4      Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Sol.

If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation.
Therefore 2 × 2 + 3 × 1 = k $\Rightarrow$ k = 4 + 3 = 7.

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