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Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 04:40

walker wrote:

Hi all! My friend, Tarek, PM me and asked me to show how to use the graphic approach to problem with inequalities. I really love such approach because it is not only fast one after training, but also reliable. So, I try to illustrate how to use it.

1) If \((x/y)>2\), is \(3x+2y<18?\)

(1) \(x-y\) is less than \(2\) (2) \(y-x\) is less than \(2\)

1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.

How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks

Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 08:24

Expert's post

kbulse wrote:

How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks

One of approaches is to check any point. Let's consider x>0 and try point between x/y=2 and x=0, for example, x=2, y=0.5. For this point x/y=4>2 and the expression is true. We could also try point above x/y=2 line. For example, x=2, y=2. For this point x/y=1 <2 and the expression is false. In conclusion, to find what area works, just pick any point in that area (the better to choose a point, for which it is possible to calculate fast). _________________

Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 09:07

walker wrote:

kbulse wrote:

How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks

One of approaches is to check any point. Let's consider x>0 and try point between x/y=2 and x=0, for example, x=2, y=0.5. For this point x/y=4>2 and the expression is true. We could also try point above x/y=2 line. For example, x=2, y=2. For this point x/y=1 <2 and the expression is false. In conclusion, to find what area works, just pick any point in that area (the better to choose a point, for which it is possible to calculate fast).

Hi walker,

how about this way: \(x/y>2\) => \(y>x/2\) then draw the line y=x/2 and shade everything which is upper side of the line? is it not correct?

Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 10:05

Expert's post

kbulse wrote:

how about this way: \(x/y>2\) => \(y>x/2\) then draw the line y=x/2 and shade everything which is upper side of the line? is it not correct?

1) you should be careful with first operation because it is an inequality and you multiply both sides by y that can have a different sign and change a sign of the inequality.

The correct answer will be: \(x/y>2\) => \(x/y * y>2 *y\) at y>=0 and \(x/y * y<2 *y\) at y<0 ==> \(y<x/2\) at y>=0 and \(y>x/2\) at y<0 _________________

Re: Graphic approach to problems with inequalities [#permalink]
11 Mar 2009, 10:12

walker wrote:

Thanks, there is a typo here: Instead of

walker wrote:

... We can put x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).

should be: ... We can put x=4.5 into x-y=2 and find that y=2.5>2.25 (left side, line x-y=2 goes above P). or even better: ... We can put y=2.25 into x-y=2 and find that x=4.25<4.5 (left side). In other words, line x-y=2 passes y=2.25 (y-coordinate of P) early and goes above P.

I've fixed it in original post. +1

Hello walker, could you please help me get the final answer? I am clear with the diagram but do not understand how to deduce the answer from it. intersection of two straight lines is (4.5,2.25). From first option: x-y<2, I am trying to confirm that whether intersection (4.5,2.25) validates this eqiation. => x-y<2 => y > x-2 => y > 4.5 - 2 => y > 2.5 But at x=4.5, actual value of y is 2.25, which is, indeed, not greater than 2.5!! What should I look for from this statement?

I guess I am missing something.....

....thank you. _________________

If You're Not Living On The Edge, You're Taking Up Too Much Space

Re: Graphic approach to problems with inequalities [#permalink]
18 Mar 2009, 19:05

Ok this might be a dumb question, but I see both graphs 4 and 5 lines passing from the true region. How can you deduce which one is correct?

Also for graph 4 "walker" said, "4. Let's consider fist condition: x-y<2. x-y=2 is a boundary. (see figure 4). As we can see all y,x that satisfies the fist condition lie in "green-TRUE" region."

But I see some points that lie outside, how can "all" points satisfy this?

I m in waiting list, gotta take the test again to get in. Please please can somebody help me ?

Re: Graphic approach to problems with inequalities [#permalink]
18 May 2009, 10:24

Expert's post

mdfrahim wrote:

Hi Walker,

This approach works for linear equation or can we use it for equations like X^2 or X^3 also ??

This approach works for any equations that could be drawn under GMAT time conditions. Y = X^3 + 1 is easy to draw but Y = 0.34X^3 - 342X^2 + 6X - 3 is not. _________________

Re: Graphic approach to problems with inequalities [#permalink]
02 Aug 2009, 09:32

Expert's post

For me this approach has one great advantage over other ones: When picture is drawn, the answer is almost obvious. Moreover, drawing is pretty straightforward process: line by line, expression by expression. Nevertheless, the more approaches you know, the more confident you are. _________________

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