EQ.1: `2x+y=4 ` and EQ.2: `y=2x`
Note that, to plot linear equations, we need at least two points. To determine the points, we may assign any values of x. Substitute it to the equation, then, solve for y.
For EQ.1 :
Value of x Substitute value of x to 2x+y=4 Value of y
0 2(0)+ y = 4 4
0 + y = 4
y = 4
2 2(2) + y = 4 0
4 + y = 4
y = 4 - 4
y = 0
So our two points for EQ.1 are (0,4) and (2,0).Plot these points and connect them. Then, extend the line on both its ends. As shown below, the graph of EQ.1 is the red line.
Value of x Substitute value of x to y=2x Value of y
0 y = 2(0) 0
y = 0
2 y = 2(2) 4
y = 4
Our two points for EQ.2 are (0,0) and (2,4). Plot these points and connect them. Then, extend the line on both ends. The resulting graph of EQ.2 is the blue line as shown below.
The solution of the sytem of equations is the intersection point of the two lines.
Base on the graph above, the intersection point is (1,2).
(A) Hence, the solution is (1,2).
To check, substitute (1,2) to the given equations. Then, determine if the resulting condition is true.
2x + y = 4
2(1) + 2 = 4
2 + 2 = 4
4 = 4 (True)
y = 2x
2 = 2(1)
2 = 2 (True)
(B) Since both conditions are true, this proves that 2x+y=4 and y=2x intersects at point (1,2).