Practice Set 1.3 Algebra 10th Standard Maths Part 1 Chapter 1 Linear Equations in Two Variables Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

10th Standard Maths 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
Fill in the blanks with correct number.
Solution:
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-1

Question 2.
Find the values of following determinants.
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-2
Solution:
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-3
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-4

Question 3.
Solve the following simultaneous equations using Cramer’s rule.
i. 3x – 4y = 10 ; 4x + 3y = 5
ii. 4x + 3y – 4 = 0 ; 6x = 8 – 5y
iii. x + 2y = -1 ; 2x – 3y = 12
iv. 6x – 4y = -12 ; 8x – 3y = -2
v. 4m + 6n = 54 ; 3m + 2n = 28
vi. 2x + 3y = 2 ; x – \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Solution:
i. The given simultaneous equations are 3x – 4y = 10 …(i)
4x + 3y = 5 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 , we get
a 1 = 3, b 1 = -4, c 1 = 10 and
a 2 = 4, b 2 = 3, c 2 = 5
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-5
∴ (x, y) = (2, -1) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are
4x + 3y – 4 = 0
∴ 4x + 3y = 4 …(i)
6x = 8 – 5y
∴ 6x + 5y = 8 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 , we get
a 1 = 4, b 1 = 3, c 1 = 4 and
a 2 = 6, b 2 = 5, c 2 = 8
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-6
∴ (x, y) = (-2, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are
x + 2y = -1 …(i)
2x – 3y = 12 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a 1 x + b 1 y = C 1 and a 2 x + b 2 y = c 2 , we get
a 1 = 1, b 1 = 2, c 1 = -1 and
a 2 = 2, b 2 = -3, c 2 = 12
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-7
∴ (x, y) = (3, -2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are
6x – 4y = -12
∴ 3x – 2y = -6 …(i) [Dividing both sides by 2]
8x – 3y = -2 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 , we get
a 1 = 3, b 1 = -2, c 1 = -6 and
a 2 = 8, b 2 = -3, c 2 = -2
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-8
∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

v. The given simultaneous equations are
4m + 6n = 54
2m + 3n = 27 …(i) [Dividing both sides by 2]
3m + 2n = 28 …(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with
a 1 m + b 1 n = c 1 and a 2 m + b 2 n = c 2 , we get
a 1 = 2, b 1 = 3, c 1 = 27 and
a 2 = 3, b 2 = 2, c 2 = 28
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-9
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-10
∴ (m, n) = (6, 5) is the solution of the given simultaneous equations.

vi. The given simultaneous equations are
2x + 3y = 2 …(i)
x = \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
∴ 2x – y = 1 …(ii) [Multiplying both sides by 2]
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 , we get
a 1 = 2, b 1 = 3, c 1 = 2 and
a 2 = 2, b 2 = -1, c 2 = 1
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-11

Question 1.
To solve the simultaneous equations by determinant method, fill in the blanks,
y + 2x – 19 = 0; 2x – 3y + 3 = 0 (Textbookpg.no. 14)
Solution:
Write the given equations in the form
ax + by = c.
2x + y = 19
2x – 3y = -3
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-12

Question 2.
Complete the following activity. (Textbook pg. no. 15)
Solution:
Maharashtra-Board-Class-10-Maths-Solutions-Chapter-1-Linear-Equations-in-Two-Variables-Practice-Set-Ex-1.3-13

Question 3.
What is the nature of solution if D = 0? (Textbook pg. no. 16)
Solution:
If D = 0, i.e. a 1 b 2 – b 1 a 2 = 0, then the two simultaneous equations do not have a unique solution.
Examples:
i. 2x – 4y = 8 and x – 2y = 4
Here, a 1 b 2 – b 1 a 2 = (2)(-2) – (-4) (1)
= -4 + 4 = 0
Graphically, we can check that these two lines coincide and hence will have infinite solutions.

ii. 2x – y = -1 and 2x – y = -4
Here, a 1 b 2 – b 1 a 2 = (2)(-1) – (-1) (2)
= -2 + 2 = 0
Graphically, we can check that these two lines are parallel and hence they do not have a solution.

Question 4.
What can you say about lines if common solution is not possible? (Textbook pg. no. 16)
Answer:
If the common solution is not possible, then the lines will either coincide or will be parallel to each other.

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