Because of all the variables here, many students are tempted to pick their own numbers to try to prove or disprove each answer choice. But an important technique for dealing with systems of inequalities involves treating them almost exactly like you would systems of equations, just with three important caveats:

1) You can only use the Elimination Method, not the Substitution Method. Since you only solve for ranges in inequalities (e.g. a < 5) and not for exact numbers (e.g. a = 5), you can't make a direct number-for-variable substitution.

2) In order to combine inequalities, the inequality signs must be pointed in the same direction.

3) When you're combining inequalities, you should always add, and never subtract.  Since subtraction of inequalities is akin to multiplying by -1 and adding, this causes errors with flipped signs and negated terms. Always look to add inequalities when you attempt to combine them.

Here, the first step is to get the signs pointing in the same direction. You already have x > r, so flip the other inequality to get s > y (which is the same thing − you’re not actually manipulating it; if y is less than s, then of course s is greater than y). Now you have:

Systems of inequalities can be solved just like systems of equations, but with three important caveats:

1) You can only use the Elimination Method, not the Substitution Method. Since you only solve for ranges in inequalities (e.g. a < 5) and not for exact numbers (e.g. a = 5), you can't make a direct number-for-variable substitution.

2) In order to combine inequalities, the inequality signs must be pointed in the same direction.

3) When you're combining inequalities, you should always add, and never subtract.  Since subtraction of inequalities is akin to multiplying by -1 and adding, this causes errors with flipped signs and negated terms. Always look to add inequalities when you attempt to combine them.

So what does that mean for you here? You have two inequalities, one dealing with  and one dealing with . But all of your answer choices are one equality with both  and  in the comparison. Since your given inequalities are both "greater than," meaning the signs are pointing in the same direction, you can add those two inequalities together:

Sums to:

And now you can just divide both sides by 3, and you have:

Which matches an answer choice and is therefore your correct answer.

Systems of inequalities can be solved just like systems of equations, but with three important caveats:

1) You can only use the Elimination Method, not the Substitution Method. Since you only solve for ranges in inequalities (e.g. a < 5) and not for exact numbers (e.g. a = 5), you can't make a direct number-for-variable substitution.

2) In order to combine inequalities, the inequality signs must be pointed in the same direction.

3) When you're combining inequalities, you should always add, and never subtract.  Since subtraction of inequalities is akin to multiplying by -1 and adding, this causes errors with flipped signs and negated terms. Always look to add inequalities when you attempt to combine them.

Here you have the signs pointing in the same direction, but you don't have the same coefficients for  in order to eliminate it to be left with only  terms (which is your goal, since you're being asked to solve for a range for ). So you will want to multiply the second inequality by 3 so that the coefficients match. That yields:

When you then stack the two inequalities and sum them, you have:

+

Yields:

You can then divide both sides by 4 to get your answer:

This systems of inequalities problem rewards you for creative algebra that allows for the transitive property. You know that , and since you're being asked about  you want to get as much value out of that statement as you can. And while you don't know exactly what  is, the second inequality does tell you about . Note that algebra allows you to add (or subtract) the same thing to both sides of an inequality, so if you want to learn more about , you can just add  to both sides of that second inequality. In doing so,  you'll find that  becomes , or .

Now you have two inequalities that each involve . If  and , then by the transitive property, .

Systems of inequalities can be solved just like systems of equations, but with three important caveats:

1) You can only use the Elimination Method, not the Substitution Method. Since you only solve for ranges in inequalities (e.g. a < 5) and not for exact numbers (e.g. a = 5), you can't make a direct number-for-variable substitution.

2) In order to combine inequalities, the inequality signs must be pointed in the same direction.

3) When you're combining inequalities, you should always add, and never subtract.  Since subtraction of inequalities is akin to multiplying by -1 and adding, this causes errors with flipped signs and negated terms. Always look to add inequalities when you attempt to combine them.

Here you should see that the  terms have the same coefficient (2), meaning that if you can move them to the same side of their respective inequalities, you'll be able to combine the inequalities and eliminate the  variable. To do so, subtract  from both sides of the second inequality, making the system:

 (the first, unchanged inequality)

 (the new second inequality)

When you sum these inequalities, you're left with:

Here is where you need to remember an important rule about inequalities: if you multiply or divide by a negative, you must flip the sign. So to divide by -2 to isolate , you will have to flip the sign:

When students face abstract inequality problems, they often pick numbers to test outcomes. But that can be time-consuming and confusing - notice that with so many variables and each given inequality including subtraction, you'd have to consider the possibilities of positive and negative numbers for each, numbers that are close together vs. far apart. There are lots of options.

The more direct way to solve features performing algebra. Notice that with two steps of algebra, you can get both inequalities in the same terms, of . If you add  to both sides of  you get:

And if you add  to both sides of  you get:

If you then combine the inequalities you know that  and , so it must be true that .

In order to combine this system of inequalities, we’ll want to get our signs pointing the same direction, so that we’re able to add the inequalities. We’re also trying to solve for the range of x in the inequality, so we’ll want to be able to eliminate our other unknown, y. In order to accomplish both of these tasks in one step, we can multiply both signs of the second inequality by -2, giving us 

3x+4y > 54 (our original first inequality)

8x-4y > -10 (our new, manipulated second inequality)

We can now add the inequalities, since our signs are the same direction (and when I start with something larger and add something larger to it, the end result will universally be larger) to arrive at 

  3x+4y > 54 

+8x-4y > -10

  11x > 44

Thus, dividing by 11 gets us to 

x > 4. Only positive 5 complies with this simplified inequality.

In order to combine this system of inequalities, we’ll want to get our signs pointing the same direction, so that we’re able to add the inequalities. We’ll also want to be able to eliminate one of our variables. In order to do so, we can multiply both sides of our second equation by -2, arriving at

x+2y > 16 (our original first inequality)

6x- 2y > -2 (our new, manipulated second inequality)

Adding these inequalities gets us to

  x+2y > 16

+6x- 2y > -2

  7x > 14

Dividing this inequality by 7 gets us to

x > 2. 

Thus, the only possible value for x in the given coordinates is 3, in the coordinate set (3, 8), our correct answer. 

Note - if you encounter an example like this one in the calculator-friendly section, you can graph the system of inequalities and see which set applies. The graph will, in this case, look like:

And we can see that the point (3, 8) falls into the overlap of both inequalities. We could also test both inequalities to see if the results comply with the set of numbers, but would likely need to invest more time in such an approach. Here, drawing conclusions on the basis of x is likely the easiest no-calculator way to go!

If you're looking for a number exactly  units to either side of , you need an absolute value equation that gives you answers of  and .

One way forward is to simply solve each equation for  using the two-case method, assuming first that the expression within the absolute value signs positive and then that it is negative and solving the resulting equation for each case.

If you do so, choice  becomes: , or  and , or , which is correct.

Choice  becomes: , or , which doesn't work. 

Choice  becomes: , or , which doesn't match. 

Choice  becomes: , or , which doesn't work.

You could also quickly solve this question if you remember the general way number line absolute value problems like this one work. Each one will be in the general form , where the "Distance" referenced is the distance from the center value to one of the two values for .

So in this case the "center value" would be  and the "distance" would be , yielding an equation of .

Any time you have an absolute value equation and a question that is asking for the smallest possible value for the expression, remember that absolute value expressions can be zero but can never be less than 0. There are two ways to solve a problem like the one given. The first is by finding the value of  such that the expression within the absolute value signs is zero and then finding the closest integer to that number. The second is by inspection (simply using brute force to find the value for xx that will minimize the value given). 

If you set , and that . Since , the closest integer value is , which will yield a value of .

The other way to approach this problem is by inspection, or through brute force. Try plugging in a few values for  to see what happens.

If  then the expression is 

If , then the expression is .

If x=2x=2 then the expression becomes .

Notice that the function decreases and then increases around , indicating that (for integer values, at least)  will yield the smallest result for .