In this chapter, the knowledge of linear equations in one variable is recalled and extended to that of two variables. Any equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables. The Maths NCERT Solutions of Class 9 offers chapter-wise solutions with precise explanations of the exercises provided in the textbook. Students can easily understand the concept of linear equations of Algebra with the help of easy examples provided in these NCERT Solutions.
Exercise 4.1
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 ₹ x and that of a pen to be ₹ y)
Solution:
Let the cost of a notebook be = ₹ x
Let the cost of a pen be = ₹ y
According to the question,
The cost of a notebook is twice the cost of a pen.
i.e., cost of a notebook = 2×cost of a pen
x = 2×y
x = 2y
x-2y = 0
x-2y = 0 is the linear equation in two variables to represent the statement, ‘The cost of a notebook is twice the cost of a pen.’
2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case.
(ii) x –(y/5)–10 = 0
Solution:
The equation x –(y/5)-10 = 0 can be written as,
1x+(-1/5)y +(–10) = 0
Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0
We get,
a = 1
b = -(1/5)
c = -10
NCERT Solutions for Class 9 Maths Chapter 3: Coordinate Geometry | Updated Pattern
(iii) –2x+3y = 6
Solution:
–2x+3y = 6
Re-arranging the equation, we get,
–2x+3y–6 = 0
The equation –2x+3y–6 = 0 can be written as,
(–2)x+3y+(– 6) = 0
Now, comparing (–2)x+3y+(–6) = 0 with ax+by+c = 0
We get, a = –2
b = 3
c =-6
(iv) x = 3y
Solution:
x = 3y
Re-arranging the equation, we get,
x-3y = 0
The equation x-3y=0 can be written as,
1x+(-3)y+(0)c = 0
Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0
We get a = 1
b = -3
c =0
(v) 2x = –5y
Solution:
2x = –5y
Re-arranging the equation, we get,
2x+5y = 0
The equation 2x+5y = 0 can be written as,
2x+5y+0 = 0
Now, comparing 2x+5y+0= 0 with ax+by+c = 0
We get a = 2
b = 5
c = 0
(vi) 3x+2 = 0
Solution:
3x+2 = 0
The equation 3x+2 = 0 can be written as,
3x+0y+2 = 0
Now comparing 3x+0+2= 0 with ax+by+c = 0
We get a = 3
b = 0
c = 2
(vii) y–2 = 0
Solution:
y–2 = 0
The equation y–2 = 0 can be written as,
0x+1y+(–2) = 0
Now comparing 0x+1y+(–2) = 0with ax+by+c = 0
We get a = 0
b = 1
c = –2
(viii) 5 = 2x
Solution:
5 = 2x
Re-arranging the equation, we get,
2x = 5
i.e., 2x–5 = 0
The equation 2x–5 = 0 can be written as,
2x+0y–5 = 0
Now comparing 2x+0y–5 = 0 with ax+by+c = 0
We get a = 2
b = 0
c = -5
Exercise 4.2
1. Which one of the following options is true, and why?
y = 3x+5 has
- A unique solution
- Only two solutions
- Infinitely many solutions
Solution:
Let us substitute different values for x in the linear equation y = 3x+5
| x | 0 | 1 | 2 | …. | 100 |
| y, where y=3x+5 | 5 | 8 | 11 | …. | 305 |
From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.
Hence, (iii) infinitely many solutions is the only option true.
2. Write four solutions for each of the following equations:
(i) 2x+y = 7
Solution:
To find the four solutions of 2x+y =7, we substitute different values for x and y.
Let x = 0
Then,
2x+y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Then,
2x+y = 7
(2×1)+y = 7
2+y = 7
y = 7-2
y = 5
(1,5)
Let y = 1
Then,
2x+y = 7
(2x)+1 = 7
2x = 7-1
2x = 6
x = 6/2
x = 3
(3,1)
Let x = 2
Then,
2x+y = 7
(2×2)+y = 7
4+y = 7
y =7-4
y = 3
(2,3)
The solutions are (0, 7), (1,5), (3,1), (2,3)
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(ii) πx+y = 9
Solution:
To find the four solutions of πx+y = 9, we substitute different values for x and y.
Let x = 0
Then,
πx+y = 9
(π×0)+y = 9
y = 9
(0,9)
Let x = 1
Then,
πx +y = 9
(π×1)+y = 9
π+y = 9
y = 9-π
(1, 9-π)
Let y = 0
Then,
πx+y = 9
πx+0 = 9
πx = 9
x = 9/π
(9/π,0)
Let x = -1
Then,
πx + y = 9
(π×-1) + y = 9
-π+y = 9
y = 9+π
(-1,9+π)
The solutions are (0,9), (1,9-π), (9/π,0), (-1,9+π)
(iii) x = 4y
Solution:
To find the four solutions of x = 4y, we substitute different values for x and y.
Let x = 0
Then,
x = 4y
0 = 4y
4y= 0
y = 0/4
y = 0
(0,0)
Let x = 1
Then,
x = 4y
1 = 4y
4y = 1
y = 1/4
(1,1/4)
Let y = 4
Then,
x = 4y
x= 4×4
x = 16
(16,4)
Let y = 1
Then,
x = 4y
x = 4×1
x = 4
(4,1)
The solutions are (0,0), (1,1/4), (16,4), (4,1)
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) (√2, 4√2)
(v) (1, 1)
Solutions:
(i) (0, 2)
(x,y) = (0,2)
Here, x=0 and y=2
Substituting the values of x and y in the equation x–2y = 4, we get,
x–2y = 4
⟹ 0 – (2×2) = 4
But, -4 ≠ 4
(0, 2) is not a solution of the equation x–2y = 4
(ii) (2, 0)
(x,y) = (2, 0)
Here, x = 2 and y = 0
Substituting the values of x and y in the equation x -2y = 4, we get,
x -2y = 4
⟹ 2-(2×0) = 4
⟹ 2 -0 = 4
But, 2 ≠ 4
(2, 0) is not a solution of the equation x-2y = 4
(iii) (4, 0)
Solution:
(x,y) = (4, 0)
Here, x= 4 and y=0
Substituting the values of x and y in the equation x -2y = 4, we get,
x–2y = 4
⟹ 4 – 2×0 = 4
⟹ 4-0 = 4
⟹ 4 = 4
(4, 0) is a solution of the equation x–2y = 4
(iv) (√2,4√2)
Solution:
(x,y) = (√2,4√2)
Here, x = √2 and y = 4√2
Substituting the values of x and y in the equation x–2y = 4, we get,
x –2y = 4
⟹ √2-(2×4√2) = 4
√2-8√2 = 4
But, -7√2 ≠ 4
(√2,4√2) is not a solution of the equation x–2y = 4
(v) (1, 1)
Solution:
(x,y) = (1, 1)
Here, x= 1 and y= 1
Substituting the values of x and y in the equation x–2y = 4, we get,
x –2y = 4
⟹ 1 -(2×1) = 4
⟹ 1-2 = 4
But, -1 ≠ 4
(1, 1) is not a solution of the equation x–2y = 4
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.
Solution:
The given equation is
2x+3y = k
According to the question, x = 2 and y = 1
Now, substituting the values of x and y in the equation2x+3y = k,
We get,
(2×2)+(3×1) = k
⟹ 4+3 = k
⟹ 7 = k
k = 7
The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.