Solve the system below graphically. Indicate the method you use to plot the lines and show work.y=2x 2x+y=4. Check by substituting the answer back into the original equations.
A) You need to find the intersections of each line to x and y axis.
Starting with the equation `y=2x` , you may find where the line intersects x axis putting y=0 such that:
2x = 0 => x = 0 => intersection point `(0,0)`
You also need to find where the line intersects y axis, hence, you need to put x=0, such that:
`y = 2*0 =gt y = 0 =gt` intersection point `(0,0)`
You need to have two point at least to draw the line and since the intersection to x and y axis is the same point, hence, you should select a value for x and then you may find the value of y such that:
`x = 1 =gt y = 2 =gt` next point `(1,2)`
You need to draw these points in a cartesian system of coordinates and then you need to draw a line that passes through these points.
You need to follow the same steps to draw the next given line `2x+y=4` , hence, you should start by finding where the line intersects x axis such that:
`y = 0 =gt 2x = 4 =gt x = 2 =gt (2,0)`
You need to find y intercept such that:
`x = 0 =gt y = 4 =gt (0,4)`
Since you have found x and y intercepts, you may draw the next line, using the same system of coordinates such that:
You should remember that the solution of a system of two equations of lines is given by the point of intersection of the lines, hence, you should draw the orthogonal projections of the point of intersection to x and y axis. Notice that x orthogonal projection falls in 1 and y orthogonal projection falls in 2.
Hence, the solution to the system is (1,2).
B) You should check if the coordinates of the point of intersection verify both equations such that:
`y =2x =gt 2 = 2*1 =gt 2=2`
`2x + y = 4 =gt 2*1 + 2 = 4 =gt 4=4`
Hence, substituting the coordinates of the point of intersection in both equations, they hold.
Hence, the coordinates of the point of intersection (1,2), represents the solution to the given system of equations.
In graphing linear equations, we need at least two points. To determine the points, we may use the slope and y-intercept of the given equations. To do so, express the equations to its corresponding slope-intrecept form which is
where m - is the slope and
b - is the y-intercept or point (0,b)
The first equation in the above problem is already in slope-intercept form. To clarify, we may re-write it as:
y = 2x
y = 2x + 0
So our slope is 2 and y-intercept is (0,0). To determine the second point, let's express the value of slope in fraction form which is `2/1` . Also, consider the sign of the slope. Since, we have a positive slope, the second point is 2 units up and 1 unit to the right of (0,0). Hence, our second point is (1,2).
Then, plot the two points and connect them. Extend the line on both its ends. The resulting graph of y=2x is the green line shown below.
For the second equation, express it in slope-intercept form by isolating y on the left side.
2x + y = 4
y = -2x + 4
The slope of the second equation is -2 ( in fraction form `-2/1` ) and the y-intercept is (0,4). Since the slope is negative, the second point is 2 units up and 1 unit to the left of (0,4). So, our second point is (-1,6).
Next, plot the two points. Connect them and extend the line on both its ends. The resulting graph of 2x+y=4 is the blue line in the figure below.
The solution of the two equations is the intersection point of the two lines. As shown above, the lines intesect at (1,2).
(A) Hence, the solution is (1,2).
To check, substitute (1,2) to the two equations.
y = 2x 2x + y = 4
2 = 2(1) 2(1) + 2 = 4
2 = 2 (True) 4 = 4 (True)
(B) Since point (1,2) satisfies both equations, then it proves that the solution to the given system of equation is (1,2).