## Linear Equations in Two Variables
Now , putting the value x = 0 in equation (1)
Putting the value x = 1 in equation (1)
Putting the value x = 2 in equation (1)
So, the table of the different solutions of the equation is (ii) The given equation is
Now , putting the value y = 0 in equation (1)
Putting the value y = 1 in equation (1)
Putting the value y = 2 in equation (1)
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)
Putting the value x = 1 in equation (1)
Putting the value x = 2 in equation (1)
So, the table of the different solutions of the equation is (iv) The given equation is
Now, putting the value x = 0 in equation (1)
Putting the value x = 1 in equation (1)
Putting the value x = 2 in equation (1)
So, the table of the different solutions of the equation is
Therefore, x = 2 and y = 14 Or, One equation is
Second equation is
Third equation is
In fact we can find infinite equations because through one point infinite lines pass.
According to problem, point (3, 4) lie on it. So, putting the value x = 3 and y = 4 in equation (1)
Taxi fare for first kilometer = Rs. 8 Taxi fare for subsequent distance = Rs. 5 Total distance covered = x Total fare = y Since the fare for first kilometer = Rs. 8 According to problem, Fare for (x – 1) kilometer = 5(x-1) So, the total fare
Hence,
Now the equation is
Now, putting the value x = 0 in equation (1)
Putting the value x = 1 in equation (1)
Putting the value x = 2 in equation (1)
So, the table of the different solutions of the equation is
Therefore the equation which satisfies these solutions is the correct equation. Equation (ii)
Putting the value x = -1 and y = 1 in the equation x + y = 0 L.H.S =
Putting the value x = 0 and y = 0 L.H.S =
Putting the value x = 1 and y = -1 L.H.S =
Hence, option (ii) x + y = 0 is correct. From the given Fig. 2, the solutions of the equation are (-1, 3), (0, 2) and (2, 0) Therefore the equation which satisfies these solutions is the correct equation. Equation (i) y = -x + 2 , satisfies these solutions.
Putting the value x = -1 and y = 3 in the equation
L.H.S =
Putting the value x = 0 and y = 2 L.H.S =
Putting the value x = 2 and y = 0 L.H.S =
Hence, option (i) y = -x + 2 is correct.
According to problem, y = 5x …………….equation (1) Now , putting the value x = 0 in equation (1)
Putting the value x = 1 in equation (1)
Putting the value x = 2 in equation (1)
So, the table of the different solutions of the equation is (i) When x = 2 units (distance) Putting the value x in equation (1)
Hence, Work done = 10. (ii) When x = 0 units (distance) Putting the value x in equation (1)
Hence, Work done = 0.
According to problem, x + y = 100 …………..(1) Now , putting the value x = 0 in equation (1)
Putting the value x = 50 in equation (1)
Putting the value x = 100 in equation (1)
So, the table of the different solutions of the equation is Q8. In countries like the USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius.
(i) We have to take Celsius along x-axis and Fahrenheit along y-axis Let C be x and F be y So, the table of the different solutions of the equation is (ii) When C = 30°, (iii) When F = 95°, (iv) When C = 0, When x° F = x° C ## Exercise 4
which is in fact y = 3 It is a line parallel to x-axis at a positive distance of 3 from it. We have two solution for it. i.e. (0, 3), (1, 3).
(ii) Given equation is
Or, It is line parallel to y-axis at a negative distance we have the two points lying it, the points are A(-4.5, 0), B(-4.5, 2). |