Solving and Inconsistant System

Keisha

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Aug 5, 2007
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Solve the system by graphing:

This is an example on a worksheet.

2x+y=2
2x+y=4

the solutuions for the first part is (0,2) (1,0)
the solutions for the second part is (0,4) (2,0)

Can someone please explain to me how they got those solutuions?
 

stapel

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I only see one system of equations. What are the "first" and "second" "parts", and for what are the listed points the solutions?

Please reply with the full and exact text of and instructions for the exercise, along with a complete description of how you graphed and the conclusion to which your graph led you.

Thank you! :D

Eliz.
 

Keisha

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On the sheet it says:

Solve the system by graphing
2x+y=2
2x+y=4


Then it says.....
we can graph the two lines as before. for 2x+y=2. two solutuions are (0,2) and (1,0). For 2x+y=4, two solutuions are (0,4) and (2,0). Using these intercepts, we graph the two equations.

There's a note that says In slope-intercept form, our equations are
y=-2x+2
and
y=-2x+4

the pictured graph shows 2 lines going diagonally \\ 2y+y=12 is at the top and 2x+y=4 is at the bottom.

Then it says: Notice that the slope for each of these lines is -2, but they have different y-intercepts. This means that the lines are parallel (they wil never intersect). because the lines have no points in common, there is no ordered pair that will satisfy both equations. The system has no solution. It is inconsistent.

I apologize for not knowing how to draw the graph on the computer.
 

galactus

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Hello Keisha:

I am just telling you what the sheet told you. What is it you need?.

The points they have come form just entering in a value of x.

You could enter x=8 into the second one and get 4-2(8)=-12

The point is (8,-12). That's all they're doing.

Here's a graph of your lines. As you know, they are parallel and do not intersect. Therefore, no solutions because they have no points in common.

Do you have a graphing calculator?. If so, just solve for y and graph.

y=-2x+2 and y=-2x+4.

The -2 is the slope and the 2 and 4 are the respective y-intercepts.

 

Subhotosh Khan

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Keisha said:
On the sheet it says:

Solve the system by graphing
2x+y=2
2x+y=4


Then it says.....
we can graph the two lines as before. for 2x+y=2. two solutuions are (0,2) and (1,0). ,

For 2x+y=4, two solutuions are (0,4) and (2,0). Using these intercepts, we graph the two equations.

In the first equation,

if you put x= 0 --> you get y = 2

if you put y= 0 --> you get x = 1

Thus you get two points (0,2) and (1,0) to draw the first line.


similarly, you can get two points for the second line

There's a note that says In slope-intercept form, our equations are
y=-2x+2
and
y=-2x+4

the pictured graph shows 2 lines going diagonally \\ 2y+y=12 is at the top and 2x+y=4 is at the bottom.

Then it says: Notice that the slope for each of these lines is -2, but they have different y-intercepts. This means that the lines are parallel (they wil never intersect). because the lines have no points in common, there is no ordered pair that will satisfy both equations. The system has no solution. It is inconsistent.

I apologize for not knowing how to draw the graph on the computer.
If you know how to draw a straight-line - given the equation, the above section should be self-explanatory.
 

stapel

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Keisha said:
for 2x+y=2. two solutuions are (0,2) and (1,0). For 2x+y=4, two solutuions are (0,4) and (2,0). Using these intercepts, we graph the two equations.
It sounds as though you are not familiar with intercepts and/or how to graph straight lines. This should have been covered, in depth, well before you go to this topic! :shock:

Since it wasn't, you will need to learn this on your own. Fortunately, there are loads of great lessons available online:

. . . . .Google results for "cartesian plane" (plotting points and graphing)

. . . . .Google results for "graphing lines"

. . . . .Google results for "intercepts"

Once you've learned how to plot points, graph lines, and locate intercepts, the "solution" you were given should make a whole lot more sense! :wink:

Eliz.
 
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