10th Maths Book Back Algebra Ex 3.1

Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 3 – Algebra Ex 3.1 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.




The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.1 Answers in English.

10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Unit 3 Chapter 1 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

Chapter 3

Exercise 3.1 Algebra

1.Solve the following system of linear equations in three variables
(i) x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16
(ii) 1x – 2y + 4 = 0; 1y – 1z + 1 = 0; 2z + 3x = 14
(iii) x + 20 = 3y2 + 10 = 2z + 5 = 110 – (y + z)
Solutions:
(i) x + y + z = 5 ………….. (1)
2x – y + z = 9 …………. (2)
x – 2y + 3z = 16 …………. (3)

10th maths unit 3 book back answer

Substitute z = 4 in (4)
3x + 2(4) = 14
3x + 8 = 14
3x = 6
x = 2
Substitute x = 2, z = 4 in (1)
2 + y + 4 = 5 ⇒ y = -1
x = 2, y = -1, z = 4

(iii) x + 20 = 3y2 + 10 = 2z + 5 = 110 – (y + z)
x = 3y2 – 10 …………. (1)
2z + 5 = 110 – (y + z)
2z = 105 – y – z
y = 105 – 3z ………….. (2)
Substitute (2) in (1), x = 3152 – 9z2 – 10
= 2z + 5 – 20
∴ 315 – 9z – 20 = 4z – 30
13 z = 315 – 20 + 30
= 325
z = 32513 = 25
x + 20 = 2z + 5
x + 20 = 50 + 5
x = 35
Substitute z = 25 in (2)
y = 105 – 3z = 105 – 75 = 30
y = 30
x = 35, y = 30, z = 25
The system has unique solutions.

 

2.Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6 ; -3x – 2y + 5z = -12 ; x – 2z = 3
(ii) 2y + z = 3 (-x + 1); -x + 3y -z = -4 3x + 2y + z = – 12
(iii) y+z4 = z+x3 = x+y2; x + y + z = 27
Solution:
(i) x + 2y – z = 6 …………. (1)
-3x – 2y + 5z = -12 ……… (2)
x – 2z = 3 …………… (3)
10th unit 3 book back answer

We see that the system has an infinite number of solutions.
(ii) 2y + z = 3(-x + 1);
-x + 3y – z = -4;
3x + 2y + z = –12
2y + z + 3x = 3 ⇒ 3x + 2y + z = 3 ………….. (1)
-x + 3y – z = -4 …………. (2)
3x + 2y + z = –12 ………………. (3)

10th unit 3 book back answer

This is a contradiction. This means the system is inconsistent and has no solutions.

10th maths unit 3 book back answer

Sub. x = 3 in (4) ⇒ 5(3) – z = 0
15 – z = 0
-z = -15
z = 15
Sub, x = 3, z = 15 in (3)
x + y + z = 27
3 + y + 15 = 27
y = 27 – 18 = 9
x = 3, y = 9, z = 15
∴ The system has unique solutions.




3.Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now?
Solution:
Let Vani’s age be x
Let Vani’s father’s age be y
Let Vani’s grand father’s age be z.

10th maths unit 3 book back answer

Sub, z = 84 in (3), we get
4x – 84 = 12
4x = 96
x = 24
Sub, x = 24, z = 84 in (1) we get
24 + y + 84 = 159
y = 159 – 108
= 51
∴ Vani’s age = 24 years
Her father’s age =51 years
Her grand father’s age = 84 years.

4.The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number?
Solution:
Let the number be 100x + 10y + z.
Reversed number be 100z + 10y + x.
x + y + z = 11 …………… (1)
100z + 10y + x = 5(100x + 10y + z) + 46
100z + 10y + x = 500x + 50y + 5z + 46
499x + 40y – 95z -46 ………….. (2)
x + 2y = z
x + 2y – z = 0 ……………. (3)

10th maths unit 3 book back answer

5.There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Solution:
Let x, y and z be number of currency pieces of 5,10,20 rupees
x + y + z = 12 ………. (1)
5x + 10y + 20z = 105 ………… (2)
10x + 5y + 20z = 125 …………. (3)

10th maths unit - 3 book back answer

Sub, z = 2 in (5), we get
15y + 20 × 2 = 85
15y = 45
y = 3
Sub; y = 3, z = 2 in (1)
x + y + z = 12
x = 7
∴ The solutions are
the number of ₹ 5 are 7
the number of ₹ 10 are 3
the number of ₹ 20 are 2

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