Question

A Class-X student Rohit, was given a task to solve the pair of the equations 2x + y = 2 and 2y-x= 4 and 2y – x = 4 graphically. To draw the graph he find three solutions of each of the given equations. After plotting the points on graph paper, he draw the lines PQ and RS, representing the equations as shown in the following graph. Based on the above graph, answer the following questions:(i) The equation of PQ from the graph, is: (a) 2y -x = 4 (b) 2x + y =2 (c) 2y +x =4 (d) 2x-y =2 (ii) The equation of RS from the graph is: (a) 2y - x = 4 (b) 2x + y = 2 (c) 2y +x = 4 (d) 2x-y = 2 (iii) The solution of the pair of equation from the graph is: (a) x = 2, y = 0 (b) x = 0, y = 2 (c) x = -4, y = 3 (d) x = 1, y = 0 (iv) One of the solution of line PQ is: (a) x = 3, y = 2 (b) x = -3, y = 2 (c) x = 2, y = 3 (d) x = 2, y = -3(v) x = 2 and y = -2 is one of the solution of the line/s: (a) PQ (b) RS (c) Both PQ and RS (d) None of these ​

Answer 1

Answer:

2x−y=2 ----- (1)

−4x−y=−4 ----- (2)

From equation (1)

y=2x−2 ------ (3)

Assume the value of x = 1, 2 and put those values in equation (3)

If x=1,y=2(1)−2=0

If x=2,y=2(2)−2=2

On the above basis, the following points are:

X 1 2

Y 0 2

Now plotting (1, 0), (2, 2) and joining them, we get a straight line.

From equation (2),

−4x−y=−4

y=−4x+4 ------ (4)

Assume the value of x=1,2 and put those values in equation (3)

If x=1,y=−4(1)+4=0

If x=2,y=−4(1)+4=−2

Now table the following points:

X 1 2

Y 0 -2

Plotting (1, 0), (2, -2) and joining them, we get another straight line.

These lines intersect at the point (1, 0) and therefore the solution of the equation is x=1,y=0