Solutions Chapter 3 of Maths of 10th Class
Linear Equations in Two Variables
Exercise 3
Q1. Draw the graph of each of the following linear equation in two variables:
(i) x + y = 4 (ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y
Sol. (i) The given equation is x + y = 4 y = 4 – x …………….equation (1)
Now , putting the value x = 0 in equation (1) y = 4 – 0 = 4. So the solution is (0, 4)
Putting the value x = 1 in equation (1) y = 4 – 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1) y = 4 – 2 = 2. So the solution is (2, 2)
So, the table of the different solutions of the equation is
(ii) The given equation is x - y = 2 x = 2 + y …………….equation (1)
Now , putting the value y = 0 in equation (1) x = 2 + 0 = 2. So the solution is (2, 0)
Putting the value y = 1 in equation (1) x = 2 + 1 = 3. So the solution is (3, 1)
Putting the value y = 2 in equation (1) x = 2 + 2 = 4. So the solution is (4, 2)
So, the table of the different solutions of the equation is (iii) The given equation is y = 3x y = 3x …………….equation (1)
Now , putting the value x = 0 in equation (1) y = 3 x 0 = 0. So the solution is (0, 0)
Putting the value x = 1 in equation (1) y = 3 x 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1) y = 3 x 2 = 6. So the solution is (2, 6)
So, the table of the different solutions of the equation is (iv) The given equation is
3 = 2 x + y
2 x + y = 3 y = 3 – 2x …………….equation (1)
Now, putting the value x = 0 in equation (1) y = 3 – 2 x 0 y = 3 – 0 = 3. So the solution is (0, 3)
Putting the value x = 1 in equation (1) y = 3 – 2 x 1 y = 3 – 2 = 1. So the solution is (1, 1)
Putting the value x = 2 in equation (1) y = 3 – 2 x 2 y = 3 – 4 = -1. So the solution is (2, -1)
So, the table of the different solutions of the equation is Q2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Sol. Since the given solution is (2, 14)
Therefore, x = 2 and y = 14
Or, One equation
Exercise 3
Q1. Draw the graph of each of the following linear equation in two variables:
(i) x + y = 4 (ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y
Sol. (i) The given equation is x + y = 4 y = 4 – x …………….equation (1)
Now , putting the value x = 0 in equation (1) y = 4 – 0 = 4. So the solution is (0, 4)
Putting the value x = 1 in equation (1) y = 4 – 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1) y = 4 – 2 = 2. So the solution is (2, 2)
So, the table of the different solutions of the equation is
(ii) The given equation is x - y = 2 x = 2 + y …………….equation (1)
Now , putting the value y = 0 in equation (1) x = 2 + 0 = 2. So the solution is (2, 0)
Putting the value y = 1 in equation (1) x = 2 + 1 = 3. So the solution is (3, 1)
Putting the value y = 2 in equation (1) x = 2 + 2 = 4. So the solution is (4, 2)
So, the table of the different solutions of the equation is (iii) The given equation is y = 3x y = 3x …………….equation (1)
Now , putting the value x = 0 in equation (1) y = 3 x 0 = 0. So the solution is (0, 0)
Putting the value x = 1 in equation (1) y = 3 x 1 = 3. So the solution is (1, 3)
Putting the value x = 2 in equation (1) y = 3 x 2 = 6. So the solution is (2, 6)
So, the table of the different solutions of the equation is (iv) The given equation is
3 = 2 x + y
2 x + y = 3 y = 3 – 2x …………….equation (1)
Now, putting the value x = 0 in equation (1) y = 3 – 2 x 0 y = 3 – 0 = 3. So the solution is (0, 3)
Putting the value x = 1 in equation (1) y = 3 – 2 x 1 y = 3 – 2 = 1. So the solution is (1, 1)
Putting the value x = 2 in equation (1) y = 3 – 2 x 2 y = 3 – 4 = -1. So the solution is (2, -1)
So, the table of the different solutions of the equation is Q2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Sol. Since the given solution is (2, 14)
Therefore, x = 2 and y = 14
Or, One equation