## RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1

Other Exercises

- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

Question 1.

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 = 12

(ii) x-\(\frac { y }{ 2 }\) -5 = 0

(iii) 2x + 3y = 9.35

(iv) 3x = -7y

(v) 2x + 3 = 0

(vi) y – 5 = 0

(vii) 4 = 3x

(viii) y = \(\frac { x }{ 2 }\)

Solution:

(i) -2x + 3y = 12

⇒ -2x + 3y – 12 = 0

Here a -2, b = 3, c = – 12

(ii) x – \(\frac { y }{ 2 }\) -5 = 0

Here a = 1, b =\(\frac { 1 }{ 2 }\) ,c = -5

(iii) 2x + 3y = 9.35

⇒ 2x + 3y – 9.35 = 0

Here a = 2, b = 3, c = – 9.35

(iv) 3x = -7y

⇒ 3x + 7y + 0 = 0

Here a = 3, b = 7,c = 0

(v) 2x + 3 = 0

⇒ 2x + 0y + 3 = 0

Here a = 2, b = 0, c = 3

(vi) y-5 = 0 ⇒ ox+y-5 = 0

Here a = 0, b = 1, c = -5

(vii) 4 = 3x

⇒ 3x – 4 = 0

⇒ 3x + 0y – 4 = 0

Here a = 3, b = 0, c = -4

(Viii) y= \(\frac { x }{ 2 }\)

⇒ \(\frac { x }{ 2 }\) – y+ 0 = 0

⇒ x-2y + 0 = 0

Here a = 1, y = -2, c = 0

Question 2.

Write each of the following as an equation in two variables.

(i) 2x = -3

(ii) y = 3

(iii) 5x = \(\frac { 7 }{ 2 }\)

(iv) y =\(\frac { 3 }{ 2 }\)x

Solution:

(i) 2x = -3⇒ 2x + 3 = 0

⇒ 2x + 0y + 3 = 0

(ii) y= 3 ⇒ y-3=0

⇒ 0x+ y-3 = 0

(iii) 5x =\(\frac { 7 }{ 2 }\) ⇒ 10x = 7

⇒ 10x + 0y – 7 = 0

(iv) y=\(\frac { 3 }{ 2 }\)x⇒2y = 3x

⇒ 3x – 2y + 0 = 0

Question 3.

The cost of ball pen is ₹5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.

Solution:

Let cost of a fountain pen = ₹x

and cost of ball pen = ₹y

∴ According to the condition,

y = \(\frac { x }{ 2 }\) -5^{
}⇒ 2y = x – 10

⇒ x – 2y – 10 = 0

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1 are helpful to complete your math homework.

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