5703 Shares

How to graph 3x-y+3=0 and 2x+6y-5=0? Topic: Corresponding angles homework help
June 19, 2019 / By Deven
Question: i have a homework about finding the equation of the bisector of the obtuse angle between this lines 3x-y+3=0 and 2x+6y-5=0 the problem is i dont know how to graph or draw 3x-y+3=0 and 2x+6y-5=0 please help me how to graph or draw it plz explain it briefly Best Answers: How to graph 3x-y+3=0 and 2x+6y-5=0? Candice | 7 days ago
Take Different values of x or y and get corresponding values for y or x, and on 2D graph put these (x,y) points and join them (minimum 2 points for an equation of a line)....... For 3x-y+3=0 : x=1 then y=6, x=2 then y=9 and form a line by joining two points (1,6) and (2,9)..... Same process with second line....
👍 122 | 👎 7
Did you like the answer? How to graph 3x-y+3=0 and 2x+6y-5=0? Share with your friends

We found more questions related to the topic: Corresponding angles homework help Originally Answered: How do you graph y = -2/3x + 1 & 4x+6y+8=0?
First write the statement correctly: y = -2x/3 + 1 This has exactly one possible interpretation. The way you wrote it can be interpreted as y = (-2/3)x or y = -2/(3x) and that can not be allowed. Since the equation is already in slope-intercept form we can avoid the usual table of values. The first point of the graph is given in the equation: (0, 1). That is why we appreciate the slope-intercept form. Mark a point on your graph paper on the y axis at y=1. Now the slope is -2/3. The minus means the line goes down as you move left to right. The 2/3 means you go 3 units to the right and 2 units down. That takes you to (3, -1) so put a mark there. Now get a straightedge and draw a line through those two points and that is your graph. 4x+6y+8=0 The usual approach is to make a table of values, assuming something and calculating a value for the other one, then drawing the graph through those points. But for a straight line it's easier to convert to the slope-intercept form. 4x+6y+8=0 The rule is you can do any valid operation on both sides of an equation and it will still be equal. So subtract 4x, the subtract 8. 6y = -4x - 8 Divide by 6. y = -2x/3 - 4/3 Graph this same as above. Amaryllis
3x - y + 3 = 0 3x + 3 = y y = 3x + 3 y-intercept is 3, and slope is 3 Mark a point on the y-axis at (0, 3). Then go up 3 and over 1 to the right for the slope of 3 = 3/1. Mark another point there at (1, 6). Draw a line through the two points. 2x + 6y - 5 = 0 6y = -2x + 5 y = (-1/3)x + 5/6 y-intercept is 5/6, and slope is -1/3. Make a point on the y-axis at (0. 5/6), which is just below (0, 1). From that point, go up 5 and over 6 to the right for the slope of 5/6. Make another point there. Draw a line through those points. By the way, the two lines are perpendicular, so the bisector forms a 45° angle.
👍 40 | 👎 1 Wallis
put it into y=Mx+b form -6y=9-3x Move 3x over -6y=-3x+9 Divide by -6y to isolate y -6y/-6=-3x/-6+9/-6 y=1/2x-3/2 On grid paper -3/2 or -1.5 is your y intercept. Draw a point. Then draw another point, up 1 (.5) and over 2 (2). Draw line to connect the two points.
👍 34 | 👎 -5 Rudolph
Plot two points on the graph which satisfy the first equation(the values of 3x and y should give you the number on the right hand side).Join the two points with a line and extend it. Now plot two points on the graph which satisfy the second equation.Join the two points with a line and extend it. The point where two lines intersect is the solution to those two equations.Hope this helps. CHEERS!!!
👍 28 | 👎 -11 Originally Answered: Graph Help?
The left one that goes up and down is always the Y axis. If your graph is over a period of time, then the easiest way is to make the X axis (running from left to right) the date axis and the temperature the Y axis. You can put "0" anywhere you like, and if there are negative numbers you probably want to go from perhaps 110 degrees to -30 degrees. You always go just a bit more than the two extremes. Draw a heavier line where "0" os at so your readers will be able to understand better. If the two variables are altitude and air temp, then I suggest you use air temp on the Y axis and altitude for the X axis, as that will give you nice rising lines left to right as you go higher in elevation. Good luck.

If you have your own answer to the question corresponding angles homework help, then you can write your own version, using the form below for an extended answer.