Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set 1.4

Maharashtra state Board 10th Standard Solutions Chapter 1 Linear Equations in two Variables – Here are all the MH Board solutions for 10th Standard maths Practice Set 1.4. This solution contains questions, answers, images, explanations of the complete Practice Set 1.4 titled Linear Equations in two Variables of maths taught in 10th Standard. If you are a student of 10th Standard who is using Maharashtra state Board Textbook to study maths, then you must come across Practice Set 1.4 Linear Equations in two Variables. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete Maharashtra Board Solutions for 10th Standard maths Chapter 1 Linear Equations in two Variables in one place.

Maharashtra State Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set 1.4

Question 1.
Solve the following simultaneous equations.
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 1
Solution:
i. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 2
∴ Equations (i) and (ii) become
2p – 3q = 15 …(iii)
8p + 5q = 77 …(iv)
Multiplying equation (iii) by 4, we get
8p – 12q = 60 …(v)
Subtracting equation (v) from (iv), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 3
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 4

ii. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 5
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 6
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 7
Substituting x = 3 in equation (vi), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 8
∴ Equations (i) and (ii) become
27p + 31q = 85 …(iii)
31p + 27q = 89 …(iv)
Adding equations (iii) and (iv), we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 9
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 10
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 11

iv. The given simultaneous equations are
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 12
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 13
Substituting x = 1 in equation (vi), we get
3(1) + y = 4
∴ 3 + y = 4
∴ y = 4 – 3 = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

Question 1.
Complete the following table. (Textbook pg. no. 16)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 14

Question 2.
In the above table the equations are not linear. Can you convert the equations into linear equations? (Textbook pg. no. 17)
Answer:
Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions.

Steps for solving equations reducible to a pair of linear equations.

  • Step 1: Select suitable variables other than those which are in the equations.
  • Step 2: Replace the given variables with new variables such that the given equations become linear equations in two variables.
  • Step 3: Solve the new simultaneous equations and find the values of the new variables.
  • Step 4: By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.

Question 3.
To solve given equations fill the below boxes suitably. (Text book pg.no. 19)
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 15

Question 4.
The examples on textbook pg. no. 17 and 18 obtained by transformation are solved by elimination method. If you solve these equations by graphical method and by Cramer’s rule will you get the same answers? Solve and check it. (Textbook pg. no. 18)
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 16
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 17 Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 18
The two lines intersect at point (1,-1).
∴ p = 1 and q = -1 is the solution of the simultaneous equations 4p + q = 3 and 2p – 3q = 5.
Re substituting the values of p and q, we get
Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Practice Set Ex 1.4 19
The two lines intersect at point (0, -1).
∴ x = 0 and y = -1 is the solution of the simultaneous equations x – y = 1 and x + y = -1.
∴ (x, y) = (0, -1) is the solution of the given simultaneous equations.

Maharashtra Board Class 10 Maths Solutions

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top