Detailed Explanation with Examples: Linear Equations | Quantitative for GMAT PDF Download

What Is a Linear Equation?

While not essential for the GMAT, understanding the definition of a linear equation holds true beyond this test. Essentially, a linear equation consists of three primary components.

• one or more variables
• variable(s) raised only to the first power
• variables that are not multiplied together.

Now, let’s look at a few examples of linear equations. When looking at these examples, ensure that you see that all three criteria above are covered.

• x + y = 10
• 2n – 5 = 12
• z/7 + y = 45

Now, let’s look at a few equations that are not linear equations:

• n² = 64

Since n is raised to the second power, we do not have a linear equation.

• xy = 24

Since we have two x and y variables multiplying each other, we do not have a linear equation.

Next, let’s discuss linear equations with just one variable:

Linear Equations with One Variable

Although linear equations can have any number of variables, the most basic ones have one variable. A few examples are:

• 5 + n = 12
• 6x – 10 = 2
• 10z – 3 = 12z + 9

Take note that the equations mentioned earlier are single-variable linear equations. It's important to realize that even if an equation contains the same variable appearing multiple times, as long as it's the same variable, it remains a single-variable equation. For instance, in the third equation where 10z and 12z are present, despite the repetition, it's still a single-variable equation due to the consistency of the variable (z).

Solving a Linear Equation with One Variable

Now that we know how to recognize a single-variable equation, let’s discuss how to solve one. The key when solving any linear equation is isolating, or getting alone, the variable that you want to solve for. When the technique is correctly executed, you will always end up with an equation that looks like the following:
variable = some value
The good news is that when solving linear equations, you can follow a step-by-step method that will work for all one-variable equations. Keep in mind that as your skills advance, you may be able to combine some of the following steps. Now, let’s practice this method with an example below.
If 4x + 5 = 2x + 10, then what is the value of x?
Step 1: First, combine the non-variable terms.
To combine the non-variable terms, we subtract 5 from both sides of the equation. Doing so will give us:
4x = 2x + 5
Step 2: Second, combine the variable terms.
To combine the variable terms, we subtract 2x from both sides of the equation. Doing so will give us:
2x = 5
Step 3 (if applicable): Finally, divide both sides of the equation by the coefficient (number in front) of the variable.
Since the coefficient of 2x is 2, we divide both sides of the equation by 2. Doing so gives us:
x = 5/2

Key Fact: To solve a linear equation with one variable, first combine the non-variable terms, then combine the variable terms. Then, if necessary, divide both sides of the equation by the coefficient of the variable.

Examples

Example 1: Linear Equation with One Variable
If 6y – 2 = 4y + 10, then y is equal to which of the following?
(a) 6
(b) 4
(c) 3
(d) 0
(e) -2
Ans: (a)
Explanation: Step 1: First, combine the non-variable terms.
To combine the non-variable terms, we add 2 to both sides of the equation. Doing so will give us:
6y = 4y + 12
Step 2: Next, combine the variable terms.
To combine the variable terms, we subtract 4y from both sides of the equation. Doing so will give us:
2y = 12
Step 3: Finally, divide both sides of the equation by the coefficient (number in front) of the variable.
Since the coefficient of 2y is 2, we divide both sides of the equation by 2. Doing so gives us:
y = 6

Example 2: Linear Equation with One Variable
If 5z – 8 = 8 – 3z, then z is equal to which of the following?
(a) 6
(b) 4
(c) 3
(d) 2
(e) -2
Ans: (d)
Explanation: Step 1: First, combine the non-variable terms.
To combine the non-variable terms, we add 8 to both sides of the equation. Doing so will give us:
5z = 16 – 3z
Step 2: Next, combine the variable terms.
To combine the variable terms, we add 3z to both sides of the equation. Doing so will give us:
8z = 16
Step 3: Finally, divide both sides of the equation by the coefficient (number in front) of the variable.
Since the coefficient of 8z is 8, we divide both sides of the equation by 8. Doing so gives us:
z = 2

Linear Equations with Two Variables

Now that we've covered equations involving a single variable, let's explore examples of equations with two variables. The primary distinction between these two types of linear equations is the presence of two variables in the latter. Here are a few illustrations:

• x + y = 24
• 2n – 10 = 4m
• 3z – y = 53

While these are a few examples of two-variable equations, we must keep in mind that usually, to determine the value of the variables in these equations, we would need a second equation with at least one of the same variables only, or a second equation with the same two variables. We refer to these multiple equations containing the same variables as systems of linear equations.

(Note: On the GMAT there are instances in which one equation containing two variables can be solved, even if a second equation is not provided. This situation is covered in more detail in our C-Trap blog.)

Solving Linear Equations with Two Variables

When tackling linear equations involving two variables or a system of equations, many steps mirror those used for single-variable equations. However, an additional step involves combining the equations to derive a final equation featuring just one variable. One approach to achieving this format is through the method known as "substitution."

Substitution Method with Two-Variable Equations

The way substitution works with linear equations is the same way substituting works in life! For example, remember when you had a substitute teacher in your high school math class? Your substitute teacher replaced your regular teacher.

In math, substitution means essentially the same thing: we are replacing one variable for another variable or variable expression. We should note that before we can use the substitution method, we need a variable isolated in at least one of the equations. Let’s practice substituting.

Substitution Practice 1:
Given the following linear equations, solve for the value of y.
x = 2y (equation 1)
x + y = 10 (equation 2)
Ans: Since we already have x isolated in equation 1, we can substitute 2y for x (in equation 2). After the substitution, equation 2 looks like this:
2y + y = 10 (equation 2)
Now that the equation has only one variable, we can easily solve for y by using the one-variable methods we previously learned.
3y = 10
y = 10/3

Substitution Practice 2:
Given the following linear equations, solve for the value of x.
2x – y = 12 (equation 1)
2x + 3y = 20 (equation 2)
Ans: Looking at the two equations, we should see that it’s easiest to isolate y in equation 1. We do so by adding y to both sides of the equation. We also need to subtract 12 from both sides of the equation. Equation 1 now looks like this:
2x – 12 = y (equation 1)
Since we now have y isolated, we can substitute 2x – 12 for y (in equation 2). Equation 2 now looks like this:
2x + 3(2x – 12) = 20 (equation 2)
We also can distribute 3 in the parentheses. When we do so, we must multiply both 2x and -12 by 3. So, we have:
2x + 6x – 36 = 20 (equation 2)
We now have a one-variable equation. We can solve for x by using the techniques we learned earlier:
2x + 6x – 36 = 20
8x – 36 = 20
8x = 56
x = 7

Key Fact: The substitution method allows you to express one variable in terms of the other in the first equation and then substitute that expression into the second equation.

Solving Two-Variable Linear Equations Using Substitution

Now that we’ve had ample practice with substituting a variable or expression from one equation to another, we can use this method to solve linear equations with two variables fully. However, we can’t stress enough how critical it is to work with two-variable equations efficiently. So, just ensure that you hone your process.

Example

Example: Linear Equations With Two Variables
If x + y = 12, and 2x = y, then x is equal to which of the following?
(a) 3
(b) 4
(c) 6
(d) 8
(e) 12
Ans: (b)
Explanation: This example is on the easier side simply because we already have y isolated in the second equation. Thus, we can immediately move to substitution.
We will substitute 2x for y in the first equation, giving us:
x + 2x = 12
3x = 12
x = 4

Solving Systems of Equations by Elimination

So far, we have discussed how to use substitution when solving systems of linear equations with two variables. As you may recall, the key to substitution is isolating one variable, allowing its equivalent to be substituted for another variable. However, what happens when our equation gets messy when trying to isolate a variable?
For example, let’s look at the equations 2x + 3y = 14 and 5x – 3y = 8, and attempt to isolate x in the first equation.
2x + 3y = 14
2x = 14 – 3y
x = (14 – 3y)/2
We should immediately see that substituting (14 – 3y)/2 for x in the second equation would be tedious and messy. The good news is that, when you find yourself in this position on the GMAT, you can usually use an optional technique: the elimination method. Let’s now discuss that process.

Key Fact: When you are unable to cleanly isolate a variable when dealing with two linear equations, you likely will need to use the elimination method.

Using the Elimination Method

Going back to the previous equations of 2x + 3y = 14 and 5x – 3y = 8, we can see that there is no “clean” way to isolate a variable in either equation.
However, we will be able to eliminate one of the variables by adding the equations. The process of adding equations is such that we add like terms.
For example, to add 2x + 3y = 14 and 5x – 3y = 8, we perform the following steps:

• 2x + 5x = 7x (Add the “x” terms from both equations.)
• 3y – 3y = 0 (Add the “y” terms from both equations.)
• 14 + 8 = 22 (Add the constant terms from both equations.)

After adding the equations together, we are left with:
7x + 0 = 22
7x = 22
x = 22/7
Generally, we now see that the elimination method is far superior to the substitution method in many cases.

Key Fact: The elimination method requires that we manipulate one or both linear equations such that when we combine the two manipulated equations, one of the variables is eliminated, allowing us to solve for the value of the remaining variable.

Examples

Example 1: Elimination Method
What is the value of y for the following system of equations?
5x – 7y = 12
3x + 7y = 4
(a) -2
(b) -2/7
(c) 0
(d) 1/2
(e) 2
Ans: (b)
Explanation: We see that these two equations can easily be combined by using the elimination method because equation 1 contains -7y and equation 2 contains 7y, and so they will immediately cancel when we combine like terms. Let’s do that now.
5x + 3x = 8x
-7y + 7y = 0
12 + 4 = 16
We now have a single-variable equation that we can solve:
8x = 16
x = 2
Note, however, that the question asked us to solve for the value of y. So, we have an additional step, which is to substitute the value of 2 for x in either equation, and then solve for y. Let’s use the first equation:
5x – 7y = 12
5(2) – 7y = 12
10 – 7y = 12
-7y = 2
y = -2/7

Example 2: Elimination Method
What is the value of x for the following system of equations?
4x – 2y = 7
6y = 5x + 14
(a) 7/4
(b) 4
(c) 5
(d) 13/2
(e) 7
Ans: (c)
Explanation: 4x – 2y = 7 (equation 1)
6y = 5x + 14 (equation 2)
Note that in equation 2 we need to subtract 5x from both sides of the equation, obtaining:
-5x + 6y = 14
Now we are able to use the elimination method to solve for x. If we multiply equation 1 by 3, we will be able to eliminate the y variable.
3(4x – 2y = 7) (equation 1)
-5x + 6y = 14 (equation 2)
12x – 6y = 21 (equation 1)
-5x + 6y = 14 (equation 2)
Adding like terms, we obtain:
12x – 5x = 7x
-6y + 6y = 0
21 + 14 = 35
Here is the result:
7x + 0 = 35
Finally, dividing both sides of the equation by 7, we have:
x = 5

Example 3: Elimination Method
Solve the following system of linear equations for x.
3x – 2y = 11
4x + 5y = 7
(a) -4
(b) 0
(c) 3
(d) 4
(e) 5
Ans: (c)
Explanation: 3x – 2y = 11 (equation 1)
4x + 5y = 7 (equation 2)
Since we want to solve for x, we would like to eliminate the y variable. However, doing so does not look easy, as we would have to multiply equation 1 by 2½ to perform the elimination method. While that approach would work, we can use an alternate technique that does not require working with fractions. Instead, we will use the concept of the Least Common Multiple (LCM) to eliminate the y variable. We see that the coefficients of the two y variables are 2 and 5. The LCM of 2 and 5 is 10. Thus, we will multiply equation 1 by 5 and equation 2 by 2, as follows:
5(3x – 2y = 11) (equation 1)
15x – 10y = 55 (equation 1)
2(4x + 5y = 7) (equation 2)
8x + 10y = 14 (equation 2)
We now see that the coefficient of the y variable in each equation is the LCM 10. We can now easily use the elimination method to solve for x. Combining like terms, we have:
15x + 8x = 23x
-10y + 10y = 0
55 + 14 = 69
Thus, we have:
23x = 69
x = 3

Sometimes We Have to Change the Sign of a Term Before Adding Equations

In the previous sections, we saw how easy it is to add two equations and eliminate one of the variables. However, there will be times when we think we can use the elimination method, but when we add the two equations together, we find that the variables do not cancel.
For example, let’s look at the equations 4n + 7m = 18 and 2n + 7m = 16. If we were to add these two equations together, we would have:
6n + 14m = 34
As we can see, we did not eliminate any variables! However, that outcome does not mean that the elimination method does not work. Rather, it means that if we were to multiply either equation by -1, and then add the equations, we would then be able to eliminate a variable! So, let’s do that now, and determine a value for n.
-1(4n + 7m = 18)
-4n – 7m = -18
Next, we can add our two equations. Let’s go term by term:
-4n + 2n = -2n
-7m + 7m = 0
-18 + 16 = -2
So, we have:
-2n + 0 = -2
-2n = -2
n = 1

Key Fact: If two two-variable equations each contain a variable with the same coefficient, we multiply one of the equations by -1. Then, we add the two equations, and one of the variables will be eliminated, allowing us to solve for the value of the remaining variable.

Examples

Example 1: Word Problem
Gertrude has 12 more stamps than Alice. If Gertrude and Alice have 36 combined stamps, then how many stamps does Alice have?
(a) 6
(b) 8
(c) 10
(d) 12
(e) 16
Ans: (d)
Explanation: First, we cancreate two variables:
A = the number of stamps Alice has
G = the number of stamps Gertrude has
Next, we can create two equations:
Since Gertrude has 12 more stamps than Alice, we have:
G = A + 12 (equation 1)
Since Gertrude and Alice have a combined total of 36 stamps, we have:
G + A = 36 (equation 2)
Finally, we can use the substitution method to determine the number of stamps that Alice has.
We can substitute A + 12 from equation 1 for G in equation 2. Doing so gives us the following:
A + 12 + A = 36
2A + 12 = 36
2A = 24
A = 12
Thus, Alice has 12 stamps.

Example 2: Word Problem
Pauline and Terrance both have pieces of candy. If Pauline has 3 times as many pieces of candy as Terrance, and together they have 48 pieces of candy, how many pieces of candy does Pauline have?
(a) 15
(b) 21
(c) 27
(d) 30
(e) 36
Ans: (e)
Explanation: First, we create two variables:
P = the number of pieces of candy Pauline has
T = the number of pieces of candy Terrance has
Next, we can create two equations:
Since Pauline has 3 times as many pieces of candy as Terrance, we have:
P = 3T (equation 1)
Since together they have 48 pieces of candy, we have:
P + T = 48 (equation 2)
Next, we can substitute 3T from equation 1 for P in equation 2. Doing so gives us:
3T + T = 48
4T = 48
T = 12
Finally, we see that since Terrance has 12 pieces of candy, Pauline has 3 x 12 = 36 pieces of candy.

Example 3: Word Problem
At a carnival, the concession stand sells hot dogs and drinks. Marta spends $12.00 for 4 hot dogs and 2 drinks. Damion spends $20.25 for 5 hot dogs and 6 drinks. What is the cost of a hot dog?
(a) $1.50
(b) $1.75
(c) $2.00
(d) $2.25
(e) $2.75
Ans: (d)
Explanation: First, we create two variables:
H = the cost of one hot dog
D = the cost of one drink
Next, we can create two equations, one for Marta’s purchase and one for Damion’s purchase:
Marta bought 4 hot dogs and 2 drinks, and she spent $12.00, so we have:
4H + 2D = 12.00 (equation 1)
Damion bought 5 hot dogs and 6 drinks, and he spent $20.25, so we have:
5H + 6D = 20.25 (equation 2)
Now, we see that the substitution method cannot easily be used, so we choose to use the elimination method. We could choose to find the LCM either of 4 and 5 OR of 2 and 6. Let’s use the LCM of 2 and 6, which is 6.
First, let’s multiply equation 1 by 3, obtaining:
3(4H + 2D = 12.00) (equation 1)
12H + 6D = 36.00 (equation 1)
We see that equations 1 and 2 now have the same coefficient of the variable D, 6, so the two equations are almost ready to be combined. However, since both of the coefficients of D are positive, we will not be able to use the elimination method until one of them is negative. Let’s multiply equation 1 by -1, obtaining:
-1(12H + 6D = 36.00) (equation 1)
-12H – 6D = -36.00 (equation 1)
Now, we combine like terms of the modified equation 1 (-12H – 6D = -36.00) and equation 2 (5H + 6D = 20.25):
-12H + 5H = -7H
-6D + 6D = 0
-36.00 + 20.25 = -15.75
Since we have eliminated variable D, we now have a single-variable equation, which we can easily solve for variable H:
-7H = -15.75
H = 2.25
Thus, the cost of one hot dog is $2.25.

Key Takeaways

• The GMAT quant section emphasizes linear equations in algebra. We've covered the definition of linear equations and basic single-variable equation solving.
• A GMAT system of linear equations involves two equations and two variables. Questions might prompt solving for one or both variables.
• Substitution method application relies on one equation being easily manipulated to solve for a variable in terms of another.
• The elimination method serves as an alternative when the substitution method becomes complex. It involves multiplying one or both equations by suitable constants to eliminate a variable when the resulting equations are added.
• If both equations feature a variable with the same coefficient, multiplying one equation by -1 allows eliminating a variable when the resulting equations are added.