Graphing an Inequality with a Number Line - GRE Quantitative Reasoning

Begin by solving for :

Now, this is represented by drawing an open circle at 6 and graphing upward to infinity:

To begin, you must simplify so that you "isolate" , (i.e. at least eliminate any coefficients from it). To do this, divide allof the members of the inequality by :

Now, this inequality represents all of the numbers between 13 and 32. However, it does include (hence, getting a closed circle for that value) and does not include (hence, getting an open circle for that value). Therefore, it looks like:

Since the inequality represents one range of values between two end points (both of which are included, given the sign being "less than or equal"), you know that whatever you answer, it must be convertible to the form:

Now, you know that it is impossible to get this out of the choices that have no absolute values involved in them. Therefore, the only options that make sense are the two having absolute values; however, here you should choose only the ones that have a , for only that will yield a range like this. Thus, we can try both of our options.

The wrong answer is simplified in this manner:

And you can stop right here, for you know you will never have for the left terminus.

The other option is simplified in this manner:

This is just what you need!

It is not necessary to solve this problem by multiplying terms out. Notice that between quantities A and B, the last three terms switch places for the two large numbers. as such they can be rewritten:

Quantity A:

Quantity B:

Both quantities A and B share the exact same terms, save for two:

Quantity A:

Quantity B:

From visual inspection, it is clear that B is larger.

Since is always positive, and , it follows that for all possible values.

For the A, it is possible to choose values that make the statement false, for example and .

C is always false.

Although C may approach zero for large values of and values of , it will never actually reach it for real values.

A and B, however, have x and y values which make them false:

For A, an example is

For B, would make the value zero.

The distance between two points is always positive. We calculate lx2 - x1l, which will give us the distance between the points.

|14- (-4)| = |14+4| = |18| = 18

Begin by solving for :

Now, this is represented by drawing an open circle at 6 and graphing upward to infinity:

To begin, you must simplify so that you "isolate" , (i.e. at least eliminate any coefficients from it). To do this, divide allof the members of the inequality by :

Now, this inequality represents all of the numbers between 13 and 32. However, it does include (hence, getting a closed circle for that value) and does not include (hence, getting an open circle for that value). Therefore, it looks like:

Since the inequality represents one range of values between two end points (both of which are included, given the sign being "less than or equal"), you know that whatever you answer, it must be convertible to the form:

Now, you know that it is impossible to get this out of the choices that have no absolute values involved in them. Therefore, the only options that make sense are the two having absolute values; however, here you should choose only the ones that have a , for only that will yield a range like this. Thus, we can try both of our options.

The wrong answer is simplified in this manner:

And you can stop right here, for you know you will never have for the left terminus.

The other option is simplified in this manner:

This is just what you need!

It is not necessary to solve this problem by multiplying terms out. Notice that between quantities A and B, the last three terms switch places for the two large numbers. as such they can be rewritten:

Quantity A:

Quantity B:

Both quantities A and B share the exact same terms, save for two:

Quantity A:

Quantity B:

From visual inspection, it is clear that B is larger.

Since is always positive, and , it follows that for all possible values.

For the A, it is possible to choose values that make the statement false, for example and .

C is always false.

Although C may approach zero for large values of and values of , it will never actually reach it for real values.

A and B, however, have x and y values which make them false:

For A, an example is

For B, would make the value zero.

The distance between two points is always positive. We calculate lx2 - x1l, which will give us the distance between the points.

|14- (-4)| = |14+4| = |18| = 18

Begin by solving for :

Now, this is represented by drawing an open circle at 6 and graphing upward to infinity:

To begin, you must simplify so that you "isolate" , (i.e. at least eliminate any coefficients from it). To do this, divide allof the members of the inequality by :

Now, this inequality represents all of the numbers between 13 and 32. However, it does include (hence, getting a closed circle for that value) and does not include (hence, getting an open circle for that value). Therefore, it looks like:

Since the inequality represents one range of values between two end points (both of which are included, given the sign being "less than or equal"), you know that whatever you answer, it must be convertible to the form:

Now, you know that it is impossible to get this out of the choices that have no absolute values involved in them. Therefore, the only options that make sense are the two having absolute values; however, here you should choose only the ones that have a , for only that will yield a range like this. Thus, we can try both of our options.

The wrong answer is simplified in this manner:

And you can stop right here, for you know you will never have for the left terminus.

The other option is simplified in this manner:

This is just what you need!

It is not necessary to solve this problem by multiplying terms out. Notice that between quantities A and B, the last three terms switch places for the two large numbers. as such they can be rewritten:

Quantity A:

Quantity B:

Both quantities A and B share the exact same terms, save for two:

Quantity A:

Quantity B:

From visual inspection, it is clear that B is larger.

Since is always positive, and , it follows that for all possible values.

For the A, it is possible to choose values that make the statement false, for example and .

C is always false.