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Re: graphs_Modulus....Help [#permalink]
08 Nov 2009, 16:41

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srini123 wrote:

Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was

what is the area bounded by graph|x/2| + |y/2| = 5?

I got hunderd since x=10 x=-10 y=10 y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

|\frac{x}{2}| + |\frac{y}{2}| = 5

After solving you'll get equation of four lines:

y=-10-x y=10+x y=10-x y=x-10

These four lines will also make a square, BUT in this case the diagonal will be 20 so the Area=\frac{20*20}{2}=200. Or the Side= \sqrt{200}, area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).

Re: graphs_Modulus....Help [#permalink]
27 Feb 2012, 22:42

1

This post received KUDOS

Expert's post

devinawilliam83 wrote:

Hi, Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:

VinuPriyaN wrote:

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards, Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.

Yes, it can be done by graphing. |x+y| + |x-y| = 4 can expand in four different wasy:

A. x+y+x-y = 4 --> x=2 B. x+y-x+y = 4 --> y=2 C. -x-y +x-y= 4 --> y=-2 D. -x-y-x+y=4 --> x=-2

So you can draw all these four lines x=2, x=-2, y=2, y=-2 to get a square with the side of 4:

I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).

Thanks in advance.

Because you are asked the area of the region bounded by |x+y| + |x-y| = 4. This equation gives you ONLY horizontal/vertical lines passing through points (2,0), (-2,0), (0,2) and (0,-2) such as x = 2, y = 2, x = -2, y = -2.

Note that x = 2 is the equation of a line (it is not a coordinate) which passes through point (2, 0) and is parallel to the y axis. Similarly, y = 2 is the equation of a line which is parallel to x axis and passes through the point (0, 2) and so on. I think you are taking x = 2 as a coordinate but that is not the case. A coordinate has a value for y too. x =2 is the equation of a line. It implies that x coordinate is always 2 and y can be anything. So all points lying on a line passing through x = 2 and parallel to y axis satisfy this criteria. _________________

Re: graphs_Modulus....Help [#permalink]
08 Nov 2009, 14:39

Expert's post

srini123 wrote:

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2 x=-2 y=2 y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4. _________________

Re: graphs_Modulus....Help [#permalink]
08 Nov 2009, 15:58

Bunuel wrote:

srini123 wrote:

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2 x=-2 y=2 y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.

Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was

what is the area bounded by graph|x/2| + |y/2| = 5?

I got hunderd since x=10 x=-10 y=10 y=-10

isnt the area 400 ? the answer given was 200, please explain _________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Re: graphs_Modulus....Help [#permalink]
22 May 2011, 07:38

Expert's post

VinuPriyaN wrote:

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards, Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them. _________________

Re: graphs_Modulus....Help [#permalink]
27 Feb 2012, 22:33

Hi, Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:

VinuPriyaN wrote:

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards, Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.