As in Exercise 9b.3

NCERT Solutions for Form 10 Maths Chapter 3 Exercise iii.iii contains the solutions to all the questions provided on page number 53 in the textbook. The NCERT Solutions Class x Maths contains stepwise solutions to all the Maths problems. These are very benign for the students while preparing for boards examinations.

NCERT Solutions Form x Maths Affiliate three- Pair of Linear Variables is an of import topic for the first term examinations and should exist dealt with in complete detail. Practising the exercises repeatedly will help the students score well in the first term examinations.

NCERT Solutions is the all-time guide for the students studying with detailed study textile including the important topics. Accurate solutions are provided by the subject experts here. The experts take tried to include all that is important. The students tin can refer the solutions for a better understanding of the topic.

Chapter 3 Pair of Linear Equations in Ii Variables

Download PDF for NCERT Solutions Course 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Practice 3.3

Admission other practice solutions of Grade 10 Maths Affiliate 3- Pair of Linear Equations in Two Variables

Exercise 3.i Solutions– three Questions
Exercise 3.two Solutions– 7 Questions
Do 3.4 Solutions– 2 Questions
Practice 3.5 Solutions– 4 Questions
Exercise three.6 Solutions– 2 Questions
Exercise iii.7 Solutions– 8 Questions

Access Answers of Maths NCERT Form 10 Chapter 3- Pair of Linear Equations in Two Variables Exercise three.iii

1. Solve the following pair of linear equations past the commutation method

(i) ten + y = 14

ten – y = 4

(two) s – t = three

(s/3) + (t/2) = vi

(three) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.three

0.4x + 0.5y = 2.three

(v) √2 ten+√iii y = 0

√3 x-√8 y = 0

(vi) (3x/ii) – (5y/3) = -ii

(x/iii) + (y/2) = (13/six)

Solutions:

(i) Given,

x + y = 14 and 10 – y = four are the two equations.

From 1^st equation, nosotros get,

x = 14 – y

At present, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = x

Or y = 5

By the value of y, we tin at present observe the verbal value of x;

∵ x = fourteen – y

∴ x = 14 – 5

Or ten = nine

Hence, x = 9 and y = 5.

(ii) Given,

s – t = three and (s/3) + (t/2) = six are the two equations.

From i^st equation, nosotros become,

southward = 3 + t ________________(1)

At present, substitute the value of s in second equation to become,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/half dozen = half-dozen

⇒ (vi+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

Now, substitute the value of t in equation (1)

due south = 3 + half-dozen = 9

Therefore, south = 9 and t = half-dozen.

(iii) Given,

3x – y = 3 and 9x – 3y = 9 are the two equations.

From i^st equation, we become,

x = (three+y)/three

Now, substitute the value of x in the given second equation to go,

9(3+y)/three – 3y = nine

⇒9 +3y -3y = 9

⇒ ix = 9

Therefore, y has infinite values and since, ten = (three+y) /3, so x also has infinite values.

(iv) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the 2 equations.

From one^st equation, we become,

x = (one.3- 0.3y)/0.two _________________(i)

Now, substitute the value of x in the given 2nd equation to go,

0.iv(1.3-0.3y)/0.ii + 0.5y = ii.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.three

⇒ 2.6 – 0.ane y = 2.3

⇒ 0.1 y = 0.iii

⇒ y = three

Now, substitute the value of y in equation (ane), nosotros get,

x = (1.3-0.three(3))/0.2 = (1.3-0.nine)/0.two = 0.iv/0.2 = two

Therefore, x = 2 and y = three.

(five) Given,

√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.

From ane^st equation, we get,

x = – (√3/√2)y __________________(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √viii y = 0

⇒ y = 0

Now, substitute the value of y in equation (1), we get,

ten = 0

Therefore, x = 0 and y = 0.

(six) Given,

(3x/2)-(5y/iii) = -two and (x/3) + (y/2) = thirteen/6 are the two equations.

From one^st equation, we become,

(iii/2)ten = -2 + (5y/3)

⇒ x = 2(-6+5y)/ix = (-12+10y)/9 ………………………(1)

Putting the value of ten in the given second equation to get,

((-12+10y)/ix)/3 + y/2 = 13/6

⇒y/2 = 13/six –( (-12+10y)/27 ) + y/2 = xiii/six

Now, substitute the value of y in equation (1), we become,

(3x/2) – 5(3)/3 = -two

⇒ (3x/ii) – 5 = -two

⇒ x = ii

Therefore, 10 = 2 and y = three.

ii. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of 'm' for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II), we get

x = (xi-3y)/2 ………………….(III)

Substituting the value of 10 in equation (II), we go

2(11-3y)/two – 4y = 24

11 – 3y – 4y = -24

-7y = -35

y = five……………………………………..(IV)

Putting the value of y in equation (Three), we get

x = (11-3×5)/2 = -4/two = -2

Hence, x = -ii, y = v

Also,

y = mx + 3

v = -2m +3

-2m = 2

grand = -i

Therefore the value of m is -one.

three. Form the pair of linear equations for the following problems and observe their solution by substitution method.

(i) The difference betwixt two numbers is 26 and i number is three times the other. Find them.

Solution:

Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (i)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we become

3x – x = 26

x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39.

(2) The larger of two supplementary angles exceeds the smaller past xviii degrees. Observe them.

Solution:

Allow the larger angle by ten^o and smaller angle exist y^o.

Nosotros know that the sum of ii supplementary pair of angles is e'er 180^o.

According to the question,

x + y = 180^o……………. (1)

x – y = 18^o……………..(2)

From (1), we get x = 180^o – y …………. (3)

Substituting (iii) in (ii), we get

180^o– y – y =18^o

162^o = 2y

y = 81^o ………….. (4)

Using the value of y in (3), we get

x = 180^o – 81^o

= 99^o

Hence, the angles are 99^o and 81^o.

(three) The motorbus of a cricket team buys vii bats and 6 balls for Rs.3800. Later, she buys iii bats and five balls for Rs.1750. Find the price of each bat and each brawl.

Solution:

Allow the toll a bat exist x and cost of a ball be y.

Co-ordinate to the question,

7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (Two)

From (I), we go

y = (3800-7x)/half-dozen………………..(Three)

Substituting (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/half-dozen = 1750 – 9500/iii

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/iii

⇒-17x = -8500

ten = 500 ……………………….. (IV)

Substituting the value of 10 in (III), we get

y = (3800-7 ×500)/6 = 300/half-dozen = 50

Hence, the price of a bat is Rs 500 and cost of a ball is Rs fifty.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a altitude of ten km, the accuse paid is Rs 105 and for a journey of fifteen km, the accuse paid is Rs 155. What are the fixed charges and the charge per km? How much does a person take to pay for travelling a distance of 25 km?

Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

According to the question,

x + 10y = 105 …………….. (ane)

x + 15y = 155 …………….. (ii)

From (i), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – x × x = five

Hence, fixed accuse is Rs five and per km charge = Rs 10

Charge for 25 km = ten + 25y = 5 + 250 = Rs 255

(5) A fraction becomes ix/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/half-dozen. Find the fraction.

Solution:

Permit the fraction exist x/y.

According to the question,

(x+2) /(y+2) = nine/11

11x + 22 = 9y + xviii

11x – 9y = -iv …………….. (1)

(x+three) /(y+iii) = 5/6

6x + eighteen = 5y +15

6x – 5y = -3 ………………. (2)

From (1), nosotros get x = (-four+9y)/11 …………….. (3)

Substituting the value of x in (ii), we get

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

-y = -9

y = ix ………………… (4)

Substituting the value of y in (iii), we go

x = (-4+9×ix )/11 = 7

Hence the fraction is 7/nine.

(vi) Five years hence, the age of Jacob will be three times that of his son. 5 years agone, Jacob'due south age was 7 times that of his son. What are their present ages?

Solutions:

Let the age of Jacob and his son be x and y respectively.

According to the question,

(x + v) = 3(y + 5)

ten – 3y = 10 …………………………………….. (1)

(x – 5) = vii(y – five)

x – 7y = -30 ………………………………………. (ii)

From (1), we get 10 = 3y + 10 ……………………. (3)

Substituting the value of x in (2), nosotros get

3y + ten – 7y = -xxx

-4y = -40

y = 10 ………………… (4)

Substituting the value of y in (3), we get

x = 3 10 ten + x = forty

Hence, the present age of Jacob and his son is 40 years and x years, respectively.

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Exercise iii.3 contains a total of 3 questions. The solutions to these questions are provided in the NCERT Solutions for Course 10 Maths Chapter iii- Pair of Linear Equations in Two Variables Exercise 3.3. The students can refer to this folio for the answers. The solutions with accurate steps are provided in the NCERT Solutions. The students will discover it helpful in case of any doubts.

Key Features of NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise three.3

• The questions are solved by the subject experts. Therefore, the answers provided are accurate.
• The solutions to all the questions in the textbook are provided hither.
• The solutions will assistance the students score well in the first term examinations.

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