RS AGGARWAL CLASS 9 CHAPTER 4 LINEAR EQUATIONS IN TWO VARIABLE EXERCISE 4A

 EXERCISE 4A 

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Question 1:

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
(i) 3x + 5y = 7.5
(ii) $2x-\frac{y}{5}+6=0$
(iii) 3y – 2x = 6
(iv) 4x = 5y
(v) $\frac{x}{5}-\frac{y}{6}=1$
(vi) $\sqrt{2}x+\sqrt{3}y=5$

Answer 1:

(i) 3x + 5y = 7.5
This can be expressed in the form ax + by + c = 0 as $3x+5y+\left(-7.5\right)=0$.
(ii) $2x-\frac{y}{5}+6=0$
This can be expressed in the form ax + by + c = 0 as $2x+\left(-\frac{1}{5}\right)y+6=0$
(iii) 3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as $2x+\left(-3y\right)+6=0$.
(iv) 4x = 5y
This can be expressed in the form ax + by + c = 0 as $4x-5y+0=0$
(v) $\frac{x}{5}-\frac{y}{6}=1$
This can be expressed in the form ax + by + c = 0 as $6x-5y=30$
(vi) $\sqrt{2}x+\sqrt{3}y=5$
This can be expressed in the form ax + by + c = 0 as $\sqrt{2}x+\sqrt{3}y+\left(-5\right)=0$

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Question 2:

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
(i) x = 6
(ii) 3x – y = x – 1
(iii) 2x + 9 = 0
(iv) 4y = 7
(v) x + y = 4
(vi) $\frac{x}{2}-\frac{y}{3}=\frac{1}{6}+y$

Answer 2:

(i) x = 6
In the form of ax + by + c = 0 we have $x+0y+(-6)=0$ where a = 1, b = 0 and $c=-6$.
(ii) 3x – y = x – 1
In the form of ax + by + c = 0 we have $2x+\left(-1y\right)+1=0$ where a = 2, b = $-1$ and c = 1
(iii) 2x + 9 = 0
In the form of ax + by + c = 0 we have $2x+\left(-1y\right)+1=0$ where a = 2, b = $-1$ and c = 1
(iv) 4y = 7
In the form of ax + by + c = 0 we have $0x+4y+\left(-7\right)=0$ where a = 0, b = 4 and c = $-7$
(v) x + y = 4
In the form of ax + by + c = 0 we have $1x+1y+\left(-4\right)=0$ where $a=1,b=1,c=-4$
(vi) $\frac{x}{2}-\frac{y}{3}=\frac{1}{6}+y$
In the form of ax + by + c = 0 we have $3x-2y=1+6y\Rightarrow 3x-8y+\left(-1\right)=0$ where $a=3,b=-8,c=-1$.

Question 3:

Check which of the following are the solutions of the equation 5x – 4y = 20.
(i) (4, 0)
(ii) (0, 5)
(iii) $\left(-2, \frac{5}{2}\right)$
(iv) (0, –5)
(v) $\left(2, \frac{-5}{2}\right)$

Answer 3:

The equation given is 5x – 4y = 20.
(i) (4, 0) 
Putting the value in the given equation we have 
$$\begin{gathered} \mathrm{LHS}:5\left(4\right)-4\left(0\right)=20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$
Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have 
$$\begin{gathered} \mathrm{LHS}:5\left(0\right)-4\left(5\right)=0-20=-20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$
Thus, (0, 5) is not a solution of the given equation.
(iii) $\left(-2, \frac{5}{2}\right)$
Putting the value in the given equation we have 
$$\begin{gathered} \mathrm{LHS}:5\left(-2\right)-4\left(\frac{5}{2}\right)=-10-10=-20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}\ne \mathrm{RHS} \end{gathered}$$
Thus, $\left(-2, \frac{5}{2}\right)$ is not a solution of the given equation.
(iv) (0, –5)
Putting the value in the given equation we have 
$$\begin{gathered} \mathrm{LHS}:5\left(0\right)-4\left(-5\right)=0+20=20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$
Thus, (0, –5) is a solution of the given equation.
(v) $\left(2, \frac{-5}{2}\right)$
Putting the value in the given equation we have 
$$\begin{gathered} \mathrm{LHS}:5\left(2\right)-4\left(\frac{-5}{2}\right)=10+10=20 \\ \mathrm{RHS}:20 \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$
Thus, $\left(2, \frac{-5}{2}\right)$ is a solution of the given equation.








 

Question 4:

Find five different solutions of each of the following equations:
(i) 2x – 3y = 6
(ii) $\frac{2x}{5}+\frac{3y}{10}=3$
(iii) 3y = 4x

Answer 4:

(i) 2x – 3y = 6
 

x	0	3	$-$3	$\frac{9}{2}$	2
y	$-$2	0	$-$4	1	$\frac{-2}{3}$

(ii) $\frac{2x}{5}+\frac{3y}{10}=3$

x	0	$\frac{15}{2}$	5	10	3
y	10	0	$\frac{10}{3}$	$\frac{-10}{3}$	6

(iii) 3y = 4x
 

x	3	$-$3	$-$6	6	0
y	4	$-$4	$-$8	8	0

Question 5:

If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.

Answer 5:

Given: 5x – 3y = k
Since x = 3 and y = 4 is a solution of the given equation so, it should satisfy the equation.
$$\begin{gathered} 5\left(3\right)-3\left(4\right)=k \\ \Rightarrow 15-12=k \\ \Rightarrow 3=k \end{gathered}$$ 
 

Question 6:

If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.

Answer 6:

Given: 4x – 3y + 1 = 0                                 .....(1)
x = 3k + 2 and y = 2k – 1
Putting these values in the equation (1) we get
$$\begin{gathered} 4\left(3k+2\right)-3\left(2k-1\right)+1=0 \\ \Rightarrow 12k+8-6k+3+1=0 \\ \Rightarrow 6k+12=0 \\ \Rightarrow k+2=0 \\ \Rightarrow k=-2 \end{gathered}$$

Question 7:

The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y).

Answer 7:

Let cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
$$\begin{gathered} \Rightarrow 5x=2y \\ \Rightarrow 5x-2y=0 \end{gathered}$$