Solve the pair of liner equations x-2y=0, 3x+4y=20
Answers
The given equations are
x-2y=0x−2y=0
3x+4y-20=03x+4y−20=0
Substitute x=0, in equation (1).
(0)-2y=0\Rightarrow y=0(0)−2y=0⇒y=0
Substitute x=2, in equation (1).
(2)-2y=0\Rightarrow y=1(2)−2y=0⇒y=1
It means line 1 passes through the point (0,0) and (2,1). Plot these two points on a coordinate plane and connect them by a straight line.
Substitute x=0, in equation (2).
3(0)+4y-20=0\Rightarrow y=53(0)+4y−20=0⇒y=5
Substitute x=4, in equation (2).
3(4)+4y-20=0\Rightarrow y=23(4)+4y−20=0⇒y=2
It means line 1 passes through the point (0,5) and (4,2). Plot these two points on a coordinate plane and connect them by a straight line.
From the below graph it is clear that the intersection point of both line is (4,2).
Therefore, the solution of the system of equations is (4,2)