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.

## Practice Set 1.3 Algebra 10th Std Maths Part 1 Answers 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 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

∴ (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

∴ (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

∴ (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

∴ (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

∴ (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

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

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_{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.