Graphing

SILVER

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Jan 27, 2011
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18
There's a question in my math book that i don't get
(attempt to) solve by graphing -x-2y=4
2x+4y=2
so I tried to double the top equation and then add it to the bottom but all that I ended up with was 0=10 witch does'nt make any sence.
 
That means your system has no solution. A solution is not guaranteed.

Look at the first equation.

Solve for y:

\(\displaystyle y= \frac{-1}{2}x-2\)

Look at the second equation. Solve for y:

\(\displaystyle y=\frac{-1}{2}x+\frac{1}{2}\)

The same slope. Remember what that means?. Parallel lines. Thus, no intersection. No intersection, no solution.

This is why you have the conflicting 10=0.

Also, if you multiply the first equation by -1, you get \(\displaystyle x+2y=-4\)

Divide the second by 2 and you get:

\(\displaystyle x+2y=1\)

See?. The same equation with different results on the right. x+2y can not equal -4 and 1 at the same time.

Once you get used to doing these, you can spot this sort of thing without working it out.

If you graph it as suggested, you will have parallel lines. No intersection and no solution.

Now suppose you had:

\(\displaystyle x+2y=4\)

\(\displaystyle 2x+4y=8\)

See?. The second equation is the first multiplied by 2. Since one equation is a multiple of the other, it has infinite solutions.

If you solve the top one for x and sub into the second, you end up with 8=8. This means infinite solutions.
 
galactus said:
That means your system has no solution. A solution is not guaranteed.

Look at the first equation.

Solve for y:

\(\displaystyle y= \frac{-1}{2}x-2\)

Look at the second equation. Solve for y:

\(\displaystyle y=\frac{-1}{2}x+\frac{1}{2}\)

\(\displaystyle > \ >\)The same slope. Remember what that means?. Parallel lines. \(\displaystyle < \ <\)


The same slope for the lines means either 1) parallel lines (no solution)

or 2) coincident lines (an infinite number of solutions).
 
Yes, look, I should have been more precise. Good point.
 
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