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.
Question 1.
Fill in the blanks with correct number.
Solution:

Question 2.
Find the values of following determinants.

Solution:

Question 3.
Solve the following simultaneous equations using Cramerβs rule.
i. 3x β 4y = 10 ; 4x + 3y = 5
ii. 4x + 3y β 4 = 0 ; 6x = 8 β 5y
iii. x + 2y = -1 ; 2x β 3y = 12
iv. 6x β 4y = -12 ; 8x β 3y = -2
v. 4m + 6n = 54 ; 3m + 2n = 28
vi. 2x + 3y = 2 ; x β \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Solution:
i. The given simultaneous equations are 3x β 4y = 10 β¦(i)
4x + 3y = 5 β¦(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a
1x + b
1y = c
1 and a
2x + b
2y = c
2, we get
a
1 = 3, b
1 = -4, c
1 = 10 and
a
2 = 4, b
2 = 3, c
2 = 5

β΄ (x, y) = (2, -1) is the solution of the given simultaneous equations.
ii. The given simultaneous equations are
4x + 3y β 4 = 0
β΄ 4x + 3y = 4 β¦(i)
6x = 8 β 5y
β΄ 6x + 5y = 8 β¦(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a
1x + b
1y = c
1 and a
2x + b
2y = c
2, we get
a
1 = 4, b
1 = 3, c
1 = 4 and
a
2 = 6, b
2 = 5, c
2 = 8

β΄ (x, y) = (-2, 4) is the solution of the given simultaneous equations.
iii. The given simultaneous equations are
x + 2y = -1 β¦(i)
2x β 3y = 12 β¦(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a
1x + b
1y = C
1 and a
2x + b
2y = c
2, we get
a
1 = 1, b
1 = 2, c
1 = -1 and
a
2 = 2, b
2 = -3, c
2 = 12

β΄ (x, y) = (3, -2) is the solution of the given simultaneous equations.
iv. The given simultaneous equations are
6x β 4y = -12
β΄ 3x β 2y = -6 β¦(i) [Dividing both sides by 2]
8x β 3y = -2 β¦(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a
1x + b
1y = c
1 and a
2x + b
2y = c
2, we get
a
1 = 3, b
1 = -2, c
1 = -6 and
a
2 = 8, b
2 = -3, c
2 = -2

β΄ (x, y) = (2, 6) is the solution of the given simultaneous equations.
v. The given simultaneous equations are
4m + 6n = 54
2m + 3n = 27 β¦(i) [Dividing both sides by 2]
3m + 2n = 28 β¦(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with
a
1m + b
1n = c
1 and a
2m + b
2n = c
2, we get
a
1 = 2, b
1 = 3, c
1 = 27 and
a
2 = 3, b
2 = 2, c
2 = 28

β΄ (m, n) = (6, 5) is the solution of the given simultaneous equations.
vi. The given simultaneous equations are
2x + 3y = 2 β¦(i)
x = \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
β΄ 2x β y = 1 β¦(ii) [Multiplying both sides by 2]
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a
1x + b
1y = c
1 and a
2x + b
2y = c
2, we get
a
1 = 2, b
1 = 3, c
1 = 2 and
a
2 = 2, b
2 = -1, c
2 = 1

Question 1.
To solve the simultaneous equations by determinant method, fill in the blanks,
y + 2x β 19 = 0; 2x β 3y + 3 = 0 (Textbookpg.no. 14)
Solution:
Write the given equations in the form
ax + by = c.
2x + y = 19
2x β 3y = -3

Question 2.
Complete the following activity. (Textbook pg. no. 15)
Solution:

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