- NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1
- NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1
Ex 4.1 Class 9 Maths 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 ₹ x and that of a pen to be ₹ y).
Solution:
Let the cost
of a notebook be ₹ X and the cost of a pen be ₹ Y.
According
to the question, we have
[The cost
of a notebook] = 2 × [The cost of a pen]
x = 2 × y
Or, x = 2y
Or, x – 2y
= 0
Thus, the
required linear equation is x – 2y = 0.
Ex 4.1 Class 9 Maths Question 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:
(i) 2x + 3y = 9.35¯
(ii) x – y/5 – 10 = 0
(iii) –2x + 3y = 6
(iv) x = 3y
(v) 2x = –5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
Solution:
(i) The
given linear equation is, 2x + 3y = 9.35¯
∴ (2)x + (3)y + (-9.35¯) = 0
On
comparing the above equation with ax
+ bx + c =
0, we have
a =
2, b =
3 and c = –9.35¯
(ii) The given linear equation is, x – y/5 – 10 = 0
⇒ x + (-1/5)y + (-10) = 0
On comparing the above equation with ax + by + c = 0, we have
a
= 1, b = –1/5 and
c = -10
(iii) The given linear equation is, -2x +
3y = 6
⇒ -2x + 3y – 6 = 0
⇒ (-2)x + (3)y + (-6) = 0
On comparing the above equation with ax + by + c – 0, we have
a = –2, b = 3 and c = –6
(iv) The given linear equation is, x =
3y
⇒ x – 3y = 0
⇒ (1)x + (-3)y + 0 = 0
On comparing the above equation with ax + by + c = 0, we have
a = 1, b = -3 and c = 0.
(v) The given linear equation is, 2x =
–5y
⇒ 2x + 5y = 0
⇒ (2)x + (5)y + 0 = 0
On comparing the above equation with ax + by + c = 0, we have
a = 2, b = 5 and c = 0.
(vi) The given linear equation is, 3x +
2 = 0
⇒ 3x + 2 + 0y = 0
⇒ (3)x + (0)y + (2) = 0
On comparing the above equation with ax + by + c = 0 , we have
a = 3, b = 0 and c = 2.
(vii) The given linear equation is, y – 2
= 0
⇒ (0)x + (1)y + (-2) = 0
On comparing the above equation with ax + by + c = 0, we have
a = 0, b = 1 and c = –2.
(viii) The given linear equation is, 5 =
2x
⇒ 5 – 2x = 0
⇒ -2x + 0y + 5 = 0
⇒ (-2)x + (0)y + (5) = 0
On comparing the above equation with ax + by + c = 0, we have
a = –2, b = 0 and c = 5.
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