solve by graphing
3x+2y=8
6x+4y=16
You need to graph the 2 equations, and find where they intersect. That is your solution.
Somebody is pulling your leg: These are the same lines, there is no solution.
I UNDERSTAND i NEED TO FIND WHERE THEY INTERSECT. THE QUESTION IS do how do I set these in slope ,y-intersect form I am having dificulity with graphying linear system.
look Danielle:
Equation 1
3x+2y=8
2y=-3x+ 8
y=-3/2 x + 4
Equation 2:
6x+4y=16
4y=-6x +16
y=-6/4 x + 16/4
y=-3/2 x + 4
Equation 1 and 2 are the same line. There is no single unique intersection. THere can be no unique solution.
You can if you want, but Bobpursley is correct. The two lines are the same! (you can see this by dividing second equation by 2, it's the same as the first). If you still need to graph it, you need to get it in the form of y = mx + b.
BObpursley I did figureed it out. If I were to graph it I would therefore have one line representing both equations. Im I correct 4 BEING THE Y-INTERSECT MOVE 3 UNITS OVER AND T WO UNITS UP?
No. Slope is negative: Move two over and three down.
Yes, you are correct. To graph the equation y = -3/2 x + 4, you can start by plotting the y-intercept, which is the point (0, 4). From there, you can use the slope, which is -3/2, to find additional points on the line.
To find the next point, move 2 units to the right and 3 units down from the y-intercept, which gives you the point (2, 1). You can continue this process to find more points, or you can draw a straight line through the two points you already plotted.
Since the second equation, 6x+4y=16, simplifies to the same line as the first equation, there is no unique solution to this system of equations. The graph will consist of a single line, representing both equations, and every point on that line will be a solution to the system.