Linear systems equation?

Question

How would I go about graphing the following system of equations?

3x + y = 4
2y = x - 6

If you could please give me a simple step by step explanation it would be very much appreciated.

Thank you.

Answer 1

You must put them into slope intercept form to graph them. Slope intercept form is y = mx + b.



3x + y = 4



You must isolate the y. To get it alone, you have to subtract 3x from bth sides. You should now have:



y = -3x + 4



To graph this, start with the point (0,4) on your graph because that is you y-intercept (where the line crosses the y-axis) then follow the slope of -3 (rise/run so it would be down 3, right 1) to find the other points on the line.



2y = x - 6



Again, you must isolate the y. To do it, divide both sides by 2. Since right now your slope is 1, it will be 1/2. Now you should have:



y = 1/2x - 3



To graph this, do the same thing that you did in your last graph. Start with point (0,-6) and use your slope of 1/2 (up 1, right 2)



Hope I was helpful!

Answer 2

Choose any number for x, say x=1. Plug in 1 for x in the first equation:

3(1)+y=4, or 3+y=4. Then solve for y: y=4-3=1. The point (1,1) is on the graph of the first equation. Repeat this process for another number for x, say x=2. After you substitute and solve for y, you get y=-2, so (2,-2) is another point on the graph of the first equation. Plot these two points, and draw a line through them. Repeat this process for the second equation. Let x=2 in the second equation, substitute and solve for y, getting y=-2. Plot the point (2,-2). If you let x=6, you get y=0, so plot the point (6,0). Then draw the line through these two points. The solution to the system of equations is where the two lines cross. You have to draw the lines carefully to get an accurate point of intersection.

Answer 3

First, you should get them into y=mx+b form. So: 3x+y=4 (subtract 3x from each side) --> y= -3x+4. And for the other, 2y=x-6 (divide each side by two) --> y= (1/2)x - 3. The y intercept for each is the constant (last number without a variable), and the slope is the coefficeint of x. So for the first equation, the Y intercept is (0,4), and the slope is minus 3. Make a point at (0,4) and go left one, up three, then draw another point (at (-1,7)). Connect these points for your first line. For the second equation, the Y Intercept is (0,-3) and the slope is 1/2. Draw a point at (0,-3) and go up one, right two and draw another point. Connect these two points for the second line.

Answer 4

you change the first equation to y = -3x + 4

the second one is 2y = x -6 then you multiply 2 on both side to get rid of the 2 so then its y = 1/2x - 3.



then you graph 4 on the y- axis and you go down 3 and over to the right 1



the next equation you graph -3 on the y axis and go up 1 and to the right 2

Answer 5

First change it into slope- intercept form. y=-3x+4 and y= 1/2x - 3

then put the y intercept on the graph i.e. 4 and -3

then use the slope to graph the rest. the first slope is -3x the other one is 1/2x

slope is rise over run so -3x is -3/1 which means it goes down 3 (because its negative) and goes over 1.

Answer 6

well depends what method, but u can do substituion, get 2y=x-6 into Slope-Intercept, then substitute the answer of that y=(1/2x)-3 and sub the (1/2x)-3 part into y for the other equation-----You can do the rest right? =P

Answer 7

2y=x-6

the other equation can be converted into:

6x+2y=4

2y= -6x+4



subtract the second equation form the first to get----



0= 7x + 2

so x= -2/7



then substitute to get y