Exercise 4.1 (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 4.1 Question 1.

The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be Rs $x$ and that of a pen to be Rs $y$ ).
Answer.

Let the cost of a notebook be Rs. $x$.

Let the cost of a pen be $R$ s. $y$.

We need to write a linear equation in two variables to represent the statement, "Cost of a notebook is twice the cost of a pen".

Therefore, we can conclude that the required statement will be $x=2 y$.

Ex 4.1 Question 2.

Express the following linear equations in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$ in each case:
(i) $2 x+3 y=9.3 \overline{5}$
(ii) $x-\frac{y}{5}-10=0$
(iii) $-2 x+3 y=6$
(iv) $x=3 y$
(v) $2 x=-5 y$
(vi) $3 x+2=0$
(vii) $y-2=0$
(viii) $5=2 x$

Answer.

(i) $2 x+3 y=9.3 \overline{5}$

We need to express the linear equation $2 x+3 y=9.3 \overline{5}$ in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$.
$2 x+3 y=9.3 \overline{5}$ can also be written as $2 x+3 y-9.3 \overline{5}=0$.

We need to compare the equation $2 x+3 y-9.3 \overline{5}=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=2, b=3$ and $c=-9.3 \overline{5}$
(ii) $x-\frac{y}{5}-10=0$

We need to express the linear equation $x-\frac{y}{5}-10=0$ in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$.
$x-\frac{y}{5}-10=0 \quad$ can also be written as $1 \cdot x-\frac{y}{5}-10=0$.

We need to compare the equation
$
1 \cdot x-\frac{y}{5}-10=0
$
with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that
$
a=1, b=-\frac{1}{5} \text { and } c=-10
$
(iii) $-2 x+3 y=6$

We need to express the linear equation $-2 x+3 y=6$ in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$.
$-2 x+3 y=6$ can also be written as $-2 x+3 y-6=0$.

We need to compare the equation $-2 x+3 y-6=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=-2, b=3$ and $c=-6$.
(iv) $x=3 y$

We need to express the linear equation $x=3 y$ in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$.
$x=3 y$ can also be written as $x-3 y+0=0$.

We need to compare the equation $x-3 y+0=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=1, b=-3$ and $c=0$.
(v) $2 x=-5 y$

We need to express the linear equation $2 x=-5 y$ in the form $a x+b y+c=0$ and indicate the values of $a$,

$2 x=-5 y$ can also be written as $2 x+5 y+0=0$.

We need to compare the equation $2 x+5 y+0=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=2, b=5$ and $c=0$.
(vi) $3 x+2=0$

We need to express the linear equation $3 x+2=0$ in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$.
$3 x+2=0$ can also be written as $3 x+0 \cdot y+2=0$.

We need to compare the equation $3 x+0 \cdot y+2=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=3, b=0$ and $c=2$.
NCERT Solutions for Class 9 Maths Exercise 4.1
(vii) $y-2=0$

We need to express the linear equation $y-2=0$ in the form $a x+b y+c=0$ and indicate the values of $a$, $b$ and $c$.

$y-2=0$ can also be written as $0 \cdot x+1 \cdot y-2=0$.

We need to compare the equation $0 \cdot x+1 \cdot y-2=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=0, b=1$ and $c=-2$.
(viii) $5=2 x$

We need to express the linear equation $5=2 x$ in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$.
$5=2 x$ can also be written as $-2 x+0 \cdot y+5=0$.

We need to compare the equation $-2 x+0 \cdot y+5=0$ with the general equation $a x+b y+c=0$, to get the values of $a, b$ and $c$.

Therefore, we can conclude that $a=-2, b=0$ and $c=5$.