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NCERT Class 9 Mathematics Chapter 4 Linear Equations in Two Variables
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Linear Equations in Two Variables
Chapter – 4
| Exercise 4.1 |
Q.1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Ans: Let the cost of a notebook is Rs. x and cost of a pen is Rs. y.
Therefore a linear equation in two variables (x and y) representing the given statement is x = 2y.
Q.2. Express the following linear equations in the form a x + by + c = 0 and indicate the values of a, b and c in each case.
Ans:
On comparing the coefficients of x and y and also constant terms we observe that
a=2,b=3 and c = –
Ans:
(iii) -2x+3y= 6
Ans: -2x+3y =6 in the form of
ax + by +c=0 is written as
-2x+3y-6=0 [Changing 6 to L.H.S.]
⇒ -2x+3y+(-6)=0
On comparing the coefficients of x and y and also constant terms we observe that
a=-2, b=3 and c=0
(iv) x = 3y
Ans: x=3y in the form of ax + by + c= 0 is written as
x-3y = 0 ⇒1.x+(-3)y-+0=0
On comparing the coefficients of x and y and also constant terms
we observe that a=1,b=-3 and c=0
(v) 2x = -5y
Ans: 2x=-5y in the form of ax + by + c= 0 is written
as 2x+5y =0 [Changing -5y to L.H.S.]
⇒ 2x+5y+0=0
On comparing the coefficients of x and y and also constant terms
we observe that a =2, b=5 and c= 0
(vi) 3x + 2 = 0
Ans: 3x + 2 = 0 in the form of ax + by + c = 0 is written as
3x+0.y +2 = 0
On comparing the coefficients of x and y and also constant terms
we observe that a =3, b=0 and c= 0
(vii) y – 2 = 0
Ans: y -2 = 0 in the form of ax + by + c = 0 is written as
0.x+1.y+(-2)=0
On comparing the coefficients of x and y and also constant terms
we observe that a =0, b=1 and c= -2
(viii) 5 = 2x
Ans: 5= 2x in the form of ax + by + c = 0 is written as 5-2.x=0
⇒ -2x+0.y+5=0
On comparing the coefficients of x and y and also constant terms
we observe that a =2, b = 0 and c=5
| Exercise 4.2 |
Q.1. Which one of the following statements is true and why?
y = 3x + 5 has
(i) a unique solution.
Ans: We are given the equation y=3x+5 Put x = 0, then y=5
Therefore, (0,5) is the solution of this equation. Again, put x=1 then y =8
(ii) Only two solutions.
Ans: Therefore, (1, 8) is also the solution
of this equation. Again, put x = 2
then y=11.
(iii) Infinitely many solutions.
Ans: Therefore, (2, 11) is also the solution of this equation. Again x=3, y=8 and so on.
Now, it is clear that this equation has infinitely many solutions.
Remember. [A linear equation in two variable has infinitely many solution.]
Q.2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Ans: (i) We have 2x + y = 7 put x = 0, then y = 7
Therefore, (0, 7) is the solution of this equation.
Again put x = 1, then x = 5
Therefore (1, 5) is also the solution of this equation. Again put x= 2,
then y = 3.
Therefore (2, 3) is also the solution of this equation. Again put x = 3
then y = 1
Therefore (3, 1) is also the solution of this equation.
Therefore, (0,7), (1,5), (2, 3) and (3, 1) are
the four solutions of the equation 2x + y = 7
(ii) лx + y = 9
Ans: We have given πx+=9 or,
Therefore (0,9) is the solution of this equation.
Again put x = 7 they y = – 13
Therefore (7,-13) is also the solution of this equation.
Again put x = 14, then y = – 35
Therefore (14,-35) is also the solution of this equation.
of equation πx+y=9
(iii) x+4y
Ans: We are given that x = 4y put y=0 then x=0
Therefore, (0, 0) is the solution of this equation.
Again y =1, then x=4.
Therefore (4, 1) is also part of this equation.
Again, put y = 2 then x= 8
Therefore, (8, 2) is also the solution of this equation.
Again put y = 3, x = 12
Therefore, (12, 3) is also the solution of this equation.
Hence, (0, 0), (4, 1), (8, 2) and (12, 3) are the four solutions of equation
x=4y
Q.3. Check which of the following are solutions of the equation
x-2y=4 and which are not.
(i) (0, 2)
Ans: We are given that x-2y=4
put x=0 and y = 2
then, 0-2×2=4,⇒-4=4
Here, L.H.S.≠R.H.S
Therefore, (0, 2) is not the solution of equation x-2y=4
(ii) (2, 0)
Ans: Put x=2 and y = 0
Then, 2-2×0=4⇒2=4
Here, L.H.S.≠R.H.S.
Therefore (2, 0) is not the solution of equation x-2y=4
(iii) (4,0)
Ans: Put x=2 and y = 0
Then 4-2×0=4⇒4=4
Here, L.H.S.=R.H.S.
Therefore, (4, 0) is the solution of equation x-2y=4
(iv) (√2, 4√2)
Ans: Put x=√2 and y = 4√2 in equation x-2y=4
Then √2-2×4√2=4 ⇒√2-8√2=4⇒7√2=4
Here, L.H.S.≠R.H.S.
Therefore, (√2, 4√2) is not the solution of equation x-2y=4
(v) (1, 1)
Ans: Put x=1 and y = 1 in equation x-2y=4
Then 1-2×1=4⇒1-2=4
L.H.S.≠R.H.S
Therefore, (1, 1) is not the solution of equation x – 2y = 4
2x + 2y = k.
Ans: We are given that x=2 and y = 1 is the solution of the equation
2x+3y=k
Put x=2 and y = 1 in equation 2x + 3y=k
2×2+3×1=k
4+3=k∴k=7
Hence, the value of k in equation 2x + 3y = k is 7
Q.4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Ans: Given Linear equation 2x + 3y = k.
We can find the value of k by substituting the values of x and y in the given equation.
Therefore, the value of k is 7.
