The Brief
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In the previous two sections, we’ve solved sets of two or even three linear equations. Here, we will do some examples of systems with four linear equations. The general strategy will be the same as before: we apply our method (either elimination or substitution) to our sets of equations in order to get rid of one equation. We repeat this until we’re left with just two or three equations, whereupon we know how to solve the problem. Also note that it takes n distinct equations to solve a system of n linear equations. (Two equations are distinct if you cannot obtain either one from the other).

Example (Elimination Method for 4 Equations)

    \[\begin{cases} 4a + 2b + 2c + d - e &= 8 \\ a - b + 3c + d &= 4 \\ 3a + 3b - c + d - e &= 6 \\ 3b - c + 2d - 3e &= -12 \\ a - b + c - d + e &= -1 \end{cases}\]

  1. Pick a variable we want to eliminate.
    Let’s pick a.
  2. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).
    Now, if we look at our equations, we see that equation 4 does not have any a’s in it. So no matter what constant we multiply the equation by, we will not get a term with a multiplied by some given coefficient. If we have an equation like this, we simply set it to the side and use it again later. This is because, recall, the point of this method is to eliminate all occurrences of a given variable, thereby simplifying our problem. Equation 4 has already eliminated all occurrences of a, so no further manipulation of the equation is needed.Now, we get:

        \[\begin{cases} 12a + 6b + 6c + 3d - 3e &= 24 \\ 12a - 12b + 36c + 12d &= 48 \\ 12a + 12b - 4c + 4d - 4e &= 24 \\ 12a - 12b + 12c - 12d + 12e &= -12 \end{cases}\]

    And we still have our equation 4:

        \[3b - c + 2d - 3e =& -12\]

  3. Subtract the equations from one another, thereby eliminating our variable.
    Note that this only applies to the equations that have a in them. And remember to use all of the equations that include a (but we only need to use each equation once).Then, we get:

        \[\begin{cases} 18b - 30c - 9d - 3e &= -24\\ -24b + 40c + 8d + 4e &= 24\\ -24b -16c + 16d - 16e &= 36 \\ 3b - c + 2d - 3e &= -12\end{cases}\]

  4. Now, we should have four equations with at most four different variables. Pick another variable to eliminate.
    Let’s eliminate b.
  5. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).

        \[\begin{cases} 18b - 30c - 9d - 3e &= -24\\ 18b - 30c - 6d - 3e &= -18\\ 18b +12c - 12d + 12e &= -27 \\ 18b - 6c + 12d - 18e &= -72\end{cases}\]

  6. Subtract the equations from one another, thereby eliminating our variable.

        \[\begin{cases} -3d &= -6\\ -42c + 6d -15e &= 9\\ 18c - 24d + 30e &= 45 \end{cases}\]

  7. Now, we should have three equations with at most three different variables.
    And we do.
  8. We can now solve this problem with the steps detailed in our previous post on the elimination method.

Example (Substitution Method with 4 Equations)

    \[\begin{cases} 4a + 2b + 2c + d - e &= 8 \\ a - b + 3c + d &= 4 \\ 3a + 3b - c + d - e &= 6 \\ 3b - c + 2d - 3e &= -12 \\ a - b + c - d + e &= -1 \end{cases}\]

  1. Pick a variable we want to substitute for.
    Let’s choose a.
  2. Use one of the lines of the equation to express that variable in terms of the other variables.
    Let’s use the second equation. Then, we get:

        \[a = 4 + b - 3c - d\]

  3. Substitute into the other equations.
    Since we used the second equation to get our equation for a, we now substitute for a in the first, third, fourth, and fifth equations. We then get (after simplifying):

        \[\begin{cases} 6b -10c -3d - e &= -8\\ 6b -10c -2d - e &= -6\\ 3b - c + 2d - 3e &= -12\\ -2c - 2d + e &= -5 \end{cases}\]

  4. Now, we should have four equations with at most four different variables.
    And this is true.
  5. Pick another variable to substitute for.
    Let’s pick e.
  6. Use one of the remaining lines of the equation to express that variable in terms of the other variables.
    Let’s use the first line. Then, we get

        \[e = 6b - 10c - 3d + 8\]

  7. Substitute into the other equations.
    Since we used the first equation, we need to plug in our formula for e into the second, third, and fourth equations. We then get (after simplifying):

        \[\begin{cases} d &= 2\\ -15b + 29c + 11d &= 12\\ 6b - 12c - 5d &= -13 \end{cases}\]

  8. Now, we should have three equations with at most three different variables.
    And we do.
  9. We can now solve this problem with the steps detailed in our previous post on the substitution method.

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In a previous post, we talked about solving systems of linear equations with the method of elimination. Here, we talk about the method of substitution.

Essentially, for this method we want to get an equation that expresses one variable in terms of the others. For example, suppose we start with the following system:

Example

    \[\begin{cases} x + 2y &= 3 \\ 2x - 3y &= 4 \end{cases}\]

We can re-write the first equation as:

    \[x = -2y + 3\]

Thereby giving us x in terms of y and some constants. We can now plug this into the second equation, giving us:

    \[2(-2y + 3) - 3y = 4\]

And simplifying, we get:

    \[-4y + 6 - 3y = -7y + 6 = 4\]

    \[\implies y = \frac{2}{7}\]

Now that we have the value of y, we can plug it into one of our equations to get the value of x:

    \[x + 2(\frac{2}{7}) = 3\]

    \[\implies x = \frac{17}{7}\]

In general, the steps of our method are as follows. Suppose we have some system of equations:

    \[\begin{cases} ax + by &= A \\ cx + dy &= B\end{cases}\]

where A, B are constants.

Substitution Method

  1. Pick one variable to substitute.

For our example, let’s pick x to substitute.

  1. Define that variable in terms of the other variables.

We can use either equation to define the value of x. Let’s use the first. Then, we get:

    \[x = \frac{A-by}{a}\]

  1. Substitute that variable into the remaining equation to solve for one of the variables.

Since we used the first equation to define x, we need to plug it into the second equation. Then, we get:

    \[cx + dy = \frac{A-by}{a}c + dy = B\]

    \[\implies y = \frac{B - \frac{A-by}{a}c}{d}\]

  1. Now that you have one of the variables, use the equations to find the other.

Just plug in your value for y into ax + by = A to find the value of x.

Now, as before, we can also use this method for systems of 3 equations. There, we can turn that problem into a system of 2 equations as follows.

Example

Suppose we have the following system of linear equations:

    \[\begin{cases} 3x + 4y - z &= 8 \\ x + 2y + 2z &= 4 \\ 2x - 3z &= 2 \end{cases}\]

Now, we pick a variable that we want to substitute. Suppose we want to substitute x first. The key is that we can pick any of the three equations to define x in terms of the other variables, y and z. Then, we use that equation to substitute for x in the other two equations.

Suppose we want to use equation 2 to define x. Then, we get:

    \[x = 4 - 2y - 2z\]

Then, since we used equation 2 to define x, we need to substitute for x in equations 1 and 3 (i.e. the other equations). We then get:

    \[\begin{cases} 3(4-2y-2z) + 4y - z &= 8 \\ 2(4-2y-2z) - 3z &= 2 \end{cases}\]

Simplifying, we get:

    \[\begin{cases}-2y - 7z &= -4 \\ -4y - 7z &= -6 \end{cases}\]

Which we can solve via our standard substitution methods.

Practice Problems

Solve the following systems of equations using the method of substitution.

  1.  

        \[\begin{cases} 7x - 2y &= x + 3 \\ x + 2y &= 0 \end{cases}\]

    Answer
  2.  

        \[\begin{cases} 7x - 2y &= 26\\ 2x + 2y &= 15 \end{cases}\]

    Answer

  3.  

        \[\begin{cases} x + 3y &= 15\\ x &= -y + 5 \end{cases}\]

     

    Answer
  4.  

        \[\begin{cases} 4x + 2y &= 3x - 3y + 3\\ 2x - y + 2 &= x - 2y \end{cases}\]

     

    Answer
  5.  

        \[\begin{cases} x + 4y - 3z &= 8 \\ x + 2y + 2z &= 4 \\ 2x - 3z = 7 \end{cases}\]

     

    Answer

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In this post, we will talk about how to solve a system of linear equations via the method of elimination. Now, you will not need to know both this method and the method of substitution that we discuss in the next post. You do need to be able to solve a system of linear equations, but either method suffices for that purpose. However, for some problems, one or the other method may be easier. So if you have the time, it is worth learning both.

Now suppose we have a system of equations as follows:

Example

    \[\begin{cases} x + 2y &= 20 \\ x + y &= 8 \end{cases}\]

We could try to solve this system by guessing values of x and y, and plugging them in to see if they work. But that would take a long time. Instead, note that if we subtract the second equation from the first, we get:

    \[x + 2y - (x+y) = 20 - 8\]

Or, in other words,

    \[y = 20-8 = 12\]

Now that we have found the value of y, we can plug it back in to find the value of x:

    \[x + (12) = 8\]

    \[\implies x = -4\]

So our solution is (-4, 12).

In general, the strategy for solving equations by elimination is to try and get some variables to cancel each other out. We ultimately want one line of the equation to have just one variable in it, for then it is easy to calculate the value of that variable. Once we have that value, we simply plug it back into an earlier equation to get the value of the remaining variable(s).

Let’s look at another example:

Example

    \[\begin{cases} 3x + y &= 8 \\ x + 2y &= 9 \end{cases}\]

Here, it is not useful to simply subtract the second equation from the first. Doing so yields:

    \[2x - y = -1\]

which is no more helpful than what we started with.

Instead, we need to make a decision: do we want to eliminate x or y? It does not matter too much which one you choose (at most you can save yourself a few lines of arithmetic). Let’s say we want to eliminate y. Then, we need to manipulate our equations so that the y variable will be eliminated. We see that in our second equation, y has a coefficient of 2 but in the first equation, y only has a coefficient of 1.

Subtracting these equations, then, will get us an equation that includes y. To get the y’s to cancel out, we need y to have the same coefficient in both equations. So let’s multiply the first equation by 2. (We could have also divided the second equation by 2, but fractions are more annoying to work with). Then, we get

    \[\begin{cases} 6x + 2y &= 16 \\ x + 2y &= 9 \end{cases}\]

And now, we can subtract the second equation from the first to get:

    \[5x = 7\]

    \[\implies x = \frac{7}{5}\]

Plugging this back into one of our initial equations, we get:

    \[x + 2y = \frac{7}{5} + 2y = 9\]

    \[\implies 2y = \frac{38}{7}\]

    \[\implies y = \frac{19}{7}\]

So our solution is (\frac{7}{5}, \frac{19}{7}).

In general, elimination problems will look like the following:

    \[\begin{cases} ax + by &= A \\ cx + dy &= B\end{cases}\]

where A, B are constants. Let’s lay out the steps for using the method of elimination in general:

Elimination Method

  1. Pick either x or y to eliminate.

For our example, suppose we want to eliminate x.

  1. Force x to have the same coefficient in both equations.

To do that here, we multiply the first equation by \frac{c}{a} to get:

    \[cx + \frac{bc}{a}y = \frac{cA}{a}\]

(The fact that we chose to manipulate the first equation is unimportant. We could have instead multiplied the second equation by \frac{a}{c} to get similar results.)

  1. Finally, we subtract the second equation from the first (or the first from the second, you can pick whichever is easier to calculate).

Here, we get:

    \[(d-\frac{bc}{a})y = \frac{cA}{a} - B\]

    \[\implies y = \frac{\frac{cA}{a} - B}{d-\frac{bc}{a}}\]

And while this looks rather complicated, recall that a, b, c, d, A, B are all just constants. So finding the value of y is now just a matter of plugging in some numbers into a calculator.

  1. Once we know the value of y, we can just plug it in to get the value of x.

Here, x = \frac{B-dy}{c}.

Now, we can also use the method of elimination to solve systems of 3 equations. Here, the trick is to turn our system of 3 equations into a system of 2 equations. This is done by picking some variable to eliminate, and then applying (roughly) the above method to eliminate it. Then, we apply the above steps method in order to solve the remaining system of 2 equations. Here is an example of such a problem.

Example

    \[\begin{cases} 3x + y - z &= 7 \\ 2x + 4y + 3z &= 24 \\ x - 2y + 2z &= 9\end{cases}\]

Suppose we want to eliminate the variable z. Then, we multiply our equations by constants so that z will have the same coefficient in every equation. Since we want to avoid working with fractions, let’s try to make z have the coefficient of 6 in every equation. Then, we multiply the first equation by -6, the second one by 2, and the third one by 3 to get:

    \[\begin{cases} -18x -6 y +6 z &= -42 \\ 4x + 8y + 6z &= 48 \\ 3x - 6y + 6z &= 27\end{cases}\]

Now, we need to cancel out the z’s. How do we do this? Now, as before, there are many ways to cancel out the z’s. The important thing is that we use all three equations in this process. (For an explanation of why this is important, see this post. Intuitively, if you don’t use all three equations, you are losing information). So we cannot simply subtract the second equation from the first and then subtract the first equation from the second.

In our case, we will simply subtract the second equation from the first, and then subtract the third equation from the second. This yields:

    \[\begin{cases} -22x -14y &= -100 \\ x + 14y &= 21 \end{cases}\]

And we can solve this using the normal method of elimination discussed earlier.

Like any mathematical ability, really becoming proficient with the method of elimination requires practice. Here are some problems to practice on.

Practice Problems

Solve the following systems of equations using the method of elimination

  1.  

        \[\begin{cases} 8x - y &= 2x + 6 \\ 2x + y &= 0 \end{cases}\]

    Answer
  1.  

        \[\begin{cases} 9x - 2y &= 26\\ 3x + 2y &= 13 \end{cases}\]

    Answer
  2.  

        \[\begin{cases} 2x + 3y &= 15 \\ x &= -y + 5 \end{cases}\]

     

    Answer
  3.  

        \[\begin{cases} 2x + 2y &= 3x - 3y + 8\\ 3x - y + 2 &= x - 2y \end{cases}\]

     

    Answer

 


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Sometimes, we want to multiply a number by itself several times. For example, if we are trying to guess someone’s password, we may want to know how many 5-letter sequences are possible. Following our lessons on combinatorics (see here), we know that there are 26 \times 26 \times 26 \times 26 \times 26 possibilities. But we may want an easier way to write this down. The solution is to use exponents. In general,

Rule for Positive Exponents

Using this rule, we can rewrite our earlier expression as simply 26^5. We call 26 the base and we call 5 the exponent or power. And in general, for b^n, b is the base and n is the exponent. In English, we say this is “b to the n-th power.” So we would say that 26^5 is 26 to the 5th power.

Now, we can take any number and raise it to the power of any positive integer simply by multiplying the number by itself that many times. So, we have:

We can even do this for irrational numbers, like \pi:

But we can also have negative exponents. When we take a negative exponent, we use the following formula:

Rule for Negative Exponents

    \[b^{-n} = (\frac{1}{b})^n\]

In essence, we flip the base and then take a normal exponent. Here are some examples:

    \[3^{-4} = (\frac{1}{3})^4 = \frac{1}{81}\]

    \[\frac{1}{8}^{-3} = \frac{8}{1}^3 = 8^3 = 192\]

    \[\pi^{-99} = \Big(\frac{1}{\pi}\Big)^{99}\]

Finally, if we raise anything (other than 0) to the 0th power, we get 1. In other words,

Rule for the 0th Power

If b \neq 0, then b^0 = 1.

For example:

    \[29^0 = 1\]

    \[\pi^0 = 1\]

    \[(49+28/3 + \sqrt{2} + \sqrt{3} + \sqrt{4892})^0 = 1\]

Now, there are some rules for how different things with exponents can be combined.

Multiplication and Division Rules for Exponents

    \[a^{b}  a^c = a^{b + c}\]

    \[\frac{a^b}{a^c} = a^{b - c}\]

Exponent Rule for Exponents

    \[(a^b)^c = a^{bc}\]

Practice Problems

Simplify the following expressions:

  1. 5^{20} \times 5^{27}
    Answer
  2. \frac{5^3}{5^{15}}
     
    Answer

  3. \frac{9^3}{3^2}
    Answer
  4. (4^2)^3
    Answer
  5. (28^3)^0
     
    Answer

  6. (7^{-3})^{-4} + \frac{5^3}{25}
     
    Answer


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Linear equations are a specific kind of equation that meets certain limitations. Specifically, linear equations are those where every variable (if there are any variables at all) is raised only to the first power (see here for a guide to powers/exponents). Furthermore, those variables must not be multiplied by any other variables.

Here are some examples of linear equations:

    \begin{align*} x =& 8 \\ 3x + 29 =& 6 \\ 3x - 4y =& 24 + z \\ 2 =& 8 - 6\end{align*}

Here are some examples of equations that are not linear:

    \begin{align*} x^2 =& 8 \\ xy = & 8 \\ x + y  =& z^2 \end{align*}

In subsequent posts, we will discuss two methods for solving linear equations: elimination and substitution. We will also talk about systems of linear inequalities, which occur when we replace those equality signs with inequalities like < or \geq.

For now, see if you can identify whether the following equations are linear or not:

Practice Problems:

Identify whether the following equations are linear or not

  1. \frac{x^2}{y} + 4 = z
    Answer
  2. 4x - 2 = 3
    Answer
  3. xy - xy + z = 4
    Answer

  4. x^{-1} + x = 3
    Answer
  5. \frac{1}{x+y} = 4
    Answer
  6. \frac{1}{x+y} =z
    Answer

Now, systems of linear equations are just groups of 1 or more linear equations. The solutions to the system are all the different ways of assigning values to each variable such that every equation in the group is true. Note that every solution must specify the value for each variable in our system. So, for example, in the following system:

    \[\begin{cases} 3x + y =& 9 \\ 2x + 2y =& 10\]

The response x = 2 is not a solution. However, x = 2, y = 3 is a solution. And in general, we can give our solutions in the form of ordered pairs. So we would write (2, 3) is a solution, where the first coordinate corresponds to the value of x and the second coordinate corresponds to the value of y.

The reason for requiring that solutions specify values for all variables (not just some) is that sometimes a certain value for one variable can be a solution and sometimes not, depending on the values of the other variables. Consider:

    \[x = y\]

Now, this system has many solutions. For example, x = 3, y = 3 is a solution. But we cannot simply say that x = 3 is a solution, for if y = 4, then that is no solution at all (since 3 \neq 4)! Thus, we require that a solution for a system of equations specify values for every variable in the system.

In our next post, we discuss how to solve systems of linear equations via elimination.


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In this series, we discuss how to solve systems of equations. This is an important skill for the GRE since many of the math problems will involve solving systems of equations (e.g. problems with inequalities, algebra problems, or word problems). Now, this content builds on some other mathematical ideas, like the idea of a variable and the distributive law. If you want a refresher on those ideas, see this post.

What is a “system of equations”?

A system of equations is a group of one or more equations, like this:

    \[x^2 + 6x + 9 = 0\]

Or this:

    \[\begin{cases} 4x + y =& 8 \\ 2x - y =& 4 \end{cases}\]

And we want to find the possible value(s) of x and any other variables in the problem. So, for example, in the second pair of equations we could have:

    \[x = 2, y = 0\]

And then we would get:

    \[\begin{cases} 4x + y = 4(2) +0 &= 8 \\ 2x - y = 2(2)- 0 &= 4\end{cases}\]

making both equations true. But if we had x = 3, y = 2 then we would get:

    \[\begin{cases} 4x + y = 4(3) +2 &= 8 \\ 2x - y = 2(3)- 2 &= 4\end{cases}\]

which would make the second equation true, but not the first. A solution to a system of equations is an assignment of values to every variable such that every equation in the system is true. By what we have said above, x = 2, y = 0 is a solution to the above pair of equations, whereas x = 3, y = 2 is not.

In this series, we will focus on solving two basic kinds of problems:

  1. Linear Systems of Equations: These only have variables raised to the 1st power. For example:
    • x + y = 21
    • \frac{42}{9} x + 2y + 81 = 13
  2. Quadratic Systems of Equations: These include variables raised to the 2nd power. For example:
    • x^2 = 2
    • 4x^2 + 2x + 6 = 13

What Kinds of Problems Involve This Skill?

This skill can come in handy for many different types of GRE problems, such as:

  • Quantity Comparison Problems
    Suppose x > 100. Which of the following is larger?

    x^2

    x + 9,900
  • Word Problems
    Tom and Liz are baking cookies. Tom can bake 10 cookies in an hour and Liz can bake 5 cookies in an hour. Working together, how long will it take them to bake 40 cookies?
  • Problems with Infinite Series
    Suppose we have the series: 1, ½, ¼, …
    What is its sum?
  • Problems with Inequalities
    Which values of x, y will satisfy the following system of inequalities?

        \[\begin{cases} x + 3y \ge  & 4 \\ 2x - y < & 1 \end{cases}\]

In our next post, we discuss what a linear equation is, a precursor to actually solving them


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In order to solve systems of equations, we need to understand three concepts: variables, how to manipulate equations, and the distributive law.

Variables

Variables are letters that stand in for numbers. Often, we will use the letters x, y, z as variables. We can use equations to assign a specific value to a certain variable. For example, x = 3 means that the variable x is equal to the number 3.

We use variables to capture certain relationships between quantities when we do not know the specific values involved. For example, I could tell you that my cell phone is worth twice as much as my laptop (and this is true). Without variables, we would have no way to represent this mathematically, since we do not know the actual value of either my laptop or cell phone. But with variables, we can say: x = the value of my laptop, y = the value of my cell phone. Then, we can represent this information mathematically as: 

    \[y = 2x\]

Thus, by using variables, we can capture relationships between unknown values.

Manipulating Equations

The second idea we need to understand is how to manipulate equations. Often, we are given a complicated equation like:

    \[3x + 8 = 17\]

And we want to turn it into a simpler equation, like x = 3.

Now, the key is that we are allowed to anything we like to one side of the equation provided we do it to the other side as well. So, for example, we can take our initial equation

    \[3x + 8 = 17\]

And we can subtract 8 from both sides to get:

    \[3x = 9\]

And we can divide by 3, provided we do it to both sides, to get:

    \[x = 3\]

So now we know what the value of x is, namely 3.

The reason why this works is that in an equation, you are saying that the left hand side is equal to the right hand side. In other words, that the left hand side represents the very same number as the right hand side. So in our equation,

    \[3x + 8 = 17\]

We are saying that 3x + 8 is just another way of writing the number 17. So of course we can do anything we like to both sides and the resulting numbers will still be equal, since we are doing the same thing to the same numbers.

Remember, though, that we still cannot do mathematically illicit operations, like dividing by 0, even if we do them to both sides.

Distributive Law

Finally, we need to understand how the distributive law works. It is a rule for how multiplication and addition interact and it states:

    \[a \times (b+c) = (b+c) \times a = a \times b +a \times c\]

An example may help in seeing that the distributive law is, in fact, true:

We know that 56 = 8\times 7, and we know that 8\times 7 = 8\times (3+4) because operations in parenthesis are done before any other operations. Thus, we know that

    \[56 = 8 \times 7 = 8 \times (3 + 4)\]

And by the distributive law,

    \[8\times (3+4) = 8 \times 3 + 8 \times 4 = 24 + 32 = 56\]

So it is true that 8\times (3+4) = 8\times 3 + 8\times 4 since 8\times (3+4) = 56 and 8\times 3 + 8\times 4 = 56 (and of course 56 = 56).

Now, one very important application of the distributive law is used in factoring quadratic equations (as we discuss here). We will need to apply to the distributive law to:

    \[(a+c)\times (b+d)\]

By distributing, we get:

    \[(a+c)\times (b+d) = (a+c)\times b + (a+c)\times d\]

    \[= a\times b +c\times b + a\times d + c\times d\]

And in general, we will omit the multiplication symbol \times and simply write:

    \[(a+c)(b+d) = ab + cb + ad + cd\]

You may have also seen this formula called FOIL (First Outer Inner Last), a mnemonic meant to remind people of the four terms in our equation:

However you choose to remember it, this formula is an important part of understanding how to factor equations.


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The T-14.

It seems like no one talks about anything else. But where did this term come from and what does it actually measure?

Where the Term "T-14" Came From

The term "T-14" refers to the top fourteen law schools according to U.S. News and World Report (USNWR), a publication that specializes in rankings. But why fourteen instead of ten or twenty-three or five? Because the top fourteen schools have stayed fairly consistent according to USNWR.

Let's repeat that for emphasis: they've stayed consistent according to USNWR.

But USNWR's rankings are somewhat arbitrary. Here's how USNWR determines them:

Notice that USNWR rankings are mostly a measure of reputation. This is a little odd, because USNWR also helps perpetuate these schools' reputations.

To some extent, the rankings are a self-fulfilling prophecy: the longer you maintain a high USNWR rank, the better your reputation…and the higher your USNWR rank. No wonder they stay static at the top.

It's also worth pointing out what the rankings are not. They are not primarily a measure of employment outcomes. They are not at all a measure of how prepared you'll be to succeed. "Faculty resources" is a poor stand-in for educational quality, as you can see in a more detailed breakdown of the ranking methodology:

Let's question some of these assumptions. Does the fact that a school spends more on its instructors, library, and supporting services mean that it is better? Maybe—or maybe it's just in a more expensive location. Does the amount a school spends on its instructors and library and supporting services really matter more than twice as much as the students' employment rate at graduation? Do you, or does anyone you know, care even 0.75% about how many volumes are in the school's library?

The point isn't that these rankings are bad. The point is that rankings—any rankings—necessarily rely on assumptions that may not hold true for you.

So, back to the most important question:

Does attending a T-14 school matter?

It depends on what you're looking for.

Graduating from a highly ranked school usually makes it easier to do the following:

  • Get a job in Big Law—especially the first time around
  • Get a federal clerkship
  • Pursue legal academia

But graduating from a T-14 school does not guarantee a job, and graduating from a school of lower ranking doesn't mean you can't get a job in Big Law. Your class rank, network, and interviewing skills also matter a lot.

If you're contemplating a specific school, you should take a look at the latest ABA-disclosed employment outcomes.

📌Further Reading


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We’ve rounded up five spectacular personal statements that helped students with borderline numbers get into T-14 schools. You’ll find these examples to be as various as a typical JD class. Some essays are about a challenge, some about the evolution of the author’s intellectual or professional journey, and some about the author’s identity. The only common thread is sincerity. The authors did not write toward an imagined idea of what an admissions officer might be looking for: they reckoned honestly with formative experiences.

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Featured image: Person Holding Blue Ballpoint Pen Writing in Notebook by picjumbo.com via Pexels is free to use without attribution.

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At 7Sage, we have worked with hundreds of law school applicants from China, South Korea, and many more countries, and we have assembled the following FAQ to help international students gain admission to America’s top law schools.

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Featured image: Close-up of World Map by Pixabay via Pexels is free to use without attribution.

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