Solve the system by graphing.
3x+y=6
3x-y=0
This is how far I got:
(First line): y=-3x+6
(Second line): y=3x
I'm not sure how to graph the second line. Is the y-intercept (0,0) and the two lines meet at (1,3)?
first of all do you have a graphing calculator?
Well using my TI-83
and keying in the equations I got
(1,3) at the intersection of the 2 lines
but for the y intercept it was aproximately
(0,6)
No, I don't have a graphing calculator. Thanks for your help.
To graph the second line, y = 3x, you can start by identifying the y-intercept and slope.
The y-intercept is where the line intersects the y-axis, and it occurs when x = 0. Plugging x = 0 into the equation, we find y = 3(0) = 0. So, the y-intercept is (0, 0).
The slope of the line is the coefficient of x, which in this case is 3. The slope represents how steep the line is. In this case, since the slope is positive, the line will slant upwards to the right.
To graph the line, you can plot the y-intercept (0, 0) on the graph, and then use the slope to find additional points. For example, you can move up 3 units on the y-axis and right 1 unit on the x-axis to find the next point (1, 3). You can continue this process to plot more points and then connect them to form a straight line.
Now let's check if your point of intersection is correct. We can plug the x-coordinate value of (1) back into the first equation to find the corresponding y-coordinate:
3x + y = 6
3(1) + y = 6
3 + y = 6
y = 6 - 3
y = 3
So, the lines do intersect at (1, 3).
To verify your solution, you can compare the two equations and check if they are satisfied:
3x + y = 6
3(1) + 3 = 6
3 + 3 = 6
6 = 6
The solution satisfies both equations, confirming that (1, 3) is the correct point of intersection.
Therefore, you correctly graphed both lines, and the point of intersection is (1, 3).