Exercise 4.2 (Revised) - Chapter 4 - Linear Equations In Two Variables - Ncert Solutions class 9 - Maths

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Chapter 4 - Linear Equations in Two Variables - NCERT Solutions Class 9 Maths

Ex 4.2 Question 1.

Which one of the following options is true, and why?
$
y=3 x+5 \text { has }
$
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions

Answer.

We need to the number of solutions of the linear equation $y=3 x+5$.

We know that any linear equation has infinitely many solutions.

Justification:
If $x=0$ then $y=3 \times 0+5=5$
If $x=1$ then $y=3 \times 1+5=8$

If $x=-2$ then $y=3 \times(-2)+5=-1$

Similarly, we can find infinite many solutions by putting the values of $x$.

Ex 4.2 Question 2.

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

Answer.

$2 x+y=7$

We know that any linear equation has infinitely many solutions.

Let us put $x=0$ in the linear equation $2 x+y=7$, to get
$
2(0)+y=7 \quad \Rightarrow y=7
$

Thus, we get first pair of solution as $(0,7)$.

Let us put $x=2$ in the linear equation $2 x+y=7$, to get
$
2(2)+y=7 \Rightarrow y+4=7 \Rightarrow y=3 \text {. }
$

Thus, we get second pair of solution as $(2,3)$.
Let us put $x=4$ in the linear equation $2 x+y=7$, to get
$
2(4)+y=7 \quad \Rightarrow y+8=7 \Rightarrow y=-1
$

Thus, we get third pair of solution as $(4,-1)$.

Let us put $x=6$ in the linear equation $2 x+y=7$, to get
$
2(6)+y=7 \Rightarrow y+12=7 \Rightarrow y=-5 \text {. }
$

Thus, we get fourth pair of solution as $(6,-5)$.
Therefore, we can conclude that four solutions for the linear equation $2 x+y=7$ are $(0,7),(2,3),(4,-1)$ and $(6,-5)$.
(ii) $\pi x+y=9$

We know that any linear equation has infinitely many solutions.
Let us put $x=0$ in the linear equation $\pi x+y=9$, to get
$
\pi(0)+y=9 \quad \Rightarrow y=9
$

Thus, we get first pair of solution as $(0,9)$.
Let us put $y=0$ in the linear equation $\pi x+y=9$, to get
$
\pi x+(0)=9 \quad \Rightarrow x=\frac{9}{\pi}
$

Thus, we get second pair of solution as $\left(\frac{9}{\pi}, 0\right)$.

Let us put $x=1$ in the linear equation $\pi x+y=9$, to get
$\pi(1)+y=9 \quad \Rightarrow y=\frac{9}{\pi}$

Thus, we get third pair of solution as $\left(1, \frac{9}{\pi}\right)$.
Let us put $y=2$ in the linear equation $\pi x+y=9$, to get
$
\pi x+2=9 \quad \Rightarrow \pi x=7 \Rightarrow x=\frac{7}{\pi}
$

Thus, we get fourth pair of solution as $\left(\frac{7}{\pi}, 2\right)$.

Therefore, we can conclude that four solutions for the linear equation $\pi x+y=9$ are $(0,9),\left(\frac{9}{\pi}, 0\right),\left(1, \frac{9}{\pi}\right)$ and $\left(\frac{7}{\pi}, 2\right)$.
(iii) $x=4 y$

We know that any linear equation has infinitely many solutions.

Let us put $y=0$ in the linear equation $x=4 y$, to get

$
x=4(0) \quad \Rightarrow x=0
$

Thus, we get first pair of solution as $(0,0)$.

Let us put $y=2$ in the linear equation $x=4 y$, to get
$
x=4(2) \quad \Rightarrow x=8
$

Thus, we get second pair of solution as $(8,2)$.

Let us put $y=4$ in the linear equation $x=4 y$, to get
$
x=4(4) \quad \Rightarrow x=16
$

Thus, we get third pair of solution as $(16,4)$.

Let us put $y=6$ in the linear equation $x=4 y$, to get
$
x=4(6) \quad \Rightarrow x=24
$

Thus, we get fourth pair of solution as $(24,6)$.

Therefore, we can conclude that four solutions for the linear equation $x=4 y$ are $(0,0),(8,2),(16,4)$ and $(24,6)$.

Ex 4.2 Question 3.

Check which of the following are solutions of the equation $x-2 y=4$ and which are not:
(i) $(0,2)$
(ii) $(2,0)$
(iii) $(4,0)$
(iv) $(\sqrt{2}, 4 \sqrt{2})$
(v) $(1,1)$

Answer.

(i) $(0,2)$
We need to put $x=0$ and $y=2$ in the L.H.S. of linear equation $x-2 y=4$, to get
$(0)-2(2)=-4$
$\therefore$ L.H.S. $\neq$ R.H.S.

Therefore, we can conclude that $(0,2)$ is not a solution of the linear equation $x-2 y=4$.
(ii) $(2,0)$

We need to put $x=2$ and $y=0$ in the L.H.S. of linear equation $x-2 y=4$, to get

$(2)-2(0)=2$
$\therefore$ L.H.S. $\neq$ R.H.S.

Therefore, we can conclude that $(2,0)$ is not a solution of the linear equation $x-2 y=4$.
(iii) $(4,0)$

We need to put $x=4$ and $y=0$ in the linear equation $x-2 y=4$, to get
(4) $-2(0)=4$
$\therefore$ L.H.S. $=$ R.H.S.

Therefore, we can conclude that $(4,0)$ is a solution of the linear equation $x-2 y=4$.
(iv) $(\sqrt{2}, 4 \sqrt{2})$

We need to put $x=\sqrt{2}$ and $y=4 \sqrt{2}$ in the linear equation $x-2 y=4$, to get
$
\begin{aligned}
& (\sqrt{2})-2(4 \sqrt{2})=-7 \sqrt{2} \\
& \therefore \text { L.H.S. } \neq \text { R.H.S. }
\end{aligned}
$

Therefore, we can conclude that $(\sqrt{2}, 4 \sqrt{2})$ is not a solution of the linear equation $x-2 y=4$.

(v) $(1,1)$

We need to put $x=1$ and $y=1$ in the linear equation $x-2 y=4$, to get
(1) $-2(1)=-1$
$\therefore$ L.H.S. $\neq$ R.H.S.

Therefore, we can conclude that ${ }^{(1,1)}$ is not a solution of the linear equation $x-2 y=4$.

Ex 4.2 Question 4.

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

Answer.

We know that, if $x=2$ and $y=1$ is a solution of the linear equation $2 x+3 y=k$, then on substituting the respective values of $x$ and $y$ in the linear equation $2 x+3 y=k$, the LHS and RHS of the given linear equation will not be effected.
$
\therefore 2(2)+3(1)=k \Rightarrow k=4+3 \Rightarrow k=7
$

Therefore, we can conclude that the value of $k$, for which the linear equation $2 x+3 y=k$ has $x=2$ and $y=1$ as one of its solutions is 7 .