SOLUTION: how do you determine linear equations and graph them?

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Question 115765: how do you determine linear equations and graph them?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Determine whether the equation is linear or not.
Then graph the equation.
If the equation, for example, is
y+=+5x+%96+3
and if we subtract 5x from both sides, then we can write the given equation as
-5x+%2B+y+=+-3

Since we can write it in the standard form,
Ax+%2B+By+=+C,
then we have a linear equation.
This means that we will have a line when we go to graph this.
1): find three ordered pair solutions
use a chart… then choose three x values
here are three x values I’m going to use:
-1,+0, and 1
(you can pick different three x values if you want)
x | y = 5x – 3 |(x, y)
-1 | y = 5(-1) - 3 = -8 |(-1, -8)
0 | y = 5(0) - 3 = -3 |(0, -3)
1 | y = 5(1) - 3 = 2|(1, 2)
2): plot the points found in step 1
3): draw the graph
here is the graph of my example:

Solved by pluggable solver: Graphing Linear Equations
In order to graph y=5%2Ax-3 we only need to plug in two points to draw the line

So lets plug in some points

Plug in x=-1

y=5%2A%28-1%29-3

y=-5-3 Multiply

y=-8 Add

So here's one point (-1,-8)




Now lets find another point

Plug in x=0

y=5%2A%280%29-3

y=0-3 Multiply

y=-3 Add

So here's another point (0,-3). Add this to our graph





Now draw a line through these points

So this is the graph of y=5%2Ax-3 through the points (-1,-8) and (0,-3)


So from the graph we can see that the slope is 5%2F1 (which tells us that in order to go from point to point we have to start at one point and go up 5 units and to the right 1 units to get to the next point) the y-intercept is (0,-3)and the x-intercept is (0.6,0) ,or (3%2F5,0)


We could graph this equation another way. Since b=-3 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,-3).


So we have one point (0,-3)





Now since the slope is 5%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,-3), we can go up 5 units



and to the right 1 units to get to our next point


Now draw a line through those points to graph y=5%2Ax-3


So this is the graph of y=5%2Ax-3 through the points (0,-3) and (1,2)