Solving Linear Equations in the GRE Math: Concepts & Examples

GRE Linear equations is a key topic in GRE Algebra and is featured in the Quant section. A linear equation (also known as a one-degree equation) is an equation in which the highest power of a variable is always 1. In mathematics, a linear equation can be put in the form,

a₁ x₁ + a₂ x₂ +........... + a_n x_n + b = 0

Where, x₁, x₂, x_n are the variables, and a₁, a₂, a_n are the coefficients, which are real numbers and b is the constant.

Linear equations can have many variables or only one. The more the variables are, the tougher the problem will be. However, most questions in the GRE Quant section are on one variable and two variables.

To solve a linear equation means to determine the values of all variables that satisfy the equation. Two equations are considered as equivalent equations if they have the same solution. For example: 2x+6=10 and 4x+12=20 are equivalent equations because both are true when x=2.

Here are some rules for forming equivalent equations:

• Adding or subtracting the same constant to both sides of an equation preserves equality, resulting in a new equation that is equivalent to the original.

• Multiplying or dividing the equation by the same non-zero constant preserves equality, resulting in a new equation that is equivalent to the original equation.

By now you have got an idea of GRE linear equations. Now let us learn how to solve linear equations or line equations in one variable and two variables with examples.

Solving Linear Equations in One Variable

The standard form of linear equation in one variable can be written in the form,

ax + b = c

Where a,b are real numbers, c is a constant and x is an unknown variable that has one solution. 

A linear equation in one variable can be solved easily by combining all the like terms on each side of the equation. Then, the rules are applied to create equivalent equations to simplify and solve. In simple terms, the variables are brought on one side, constants on the other side, to get the value of an unknown variable. Check out the examples given below to understand how to solve one variable linear equation in the GRE exam.

Example 1:

Solve 9x - 3x + 15 = 4x + 3(x-3)

Answer Explanation:

9x - 3x + 15 = 4x + 3(x-3)

6x + 15 = 4x + 3x - 9

6x + 15 = 7x - 9

Now combine the like terms on each side of the equation

15 + 9 = 7x - 6x

24 = x

Hence the solution is x = 24.

Example 2:

If bx + c = d - ax, then what is x?

Answer Explanation:

To solve for x:

bx + c = d - ax

Now combine the like terms on each side of the equation

bx + ax = d - c

x(b+a) = d - c

x = (d - c) / (b + a)

Solving Linear Equations in Two Variables

The standard form of linear equation in two variables x and y can be written in the form

ax + by + c = 0

Where a and b are real numbers but are not zero at the same time and x and y are two variables each with a degree of 1. If we consider two such linear equations, they are called simultaneous linear equations.  We can solve two variable equations by any of these methods- 1. Substitution method 2. Elimination method. Check out the examples given below to learn how to solve two variable equations in the GRE Math.

Example 1

Solve using substitution and elimination method- 3x + 3y = 6 and 6x - 4y = 18

Solution:

Method 1: Using substitution method:

This method involves solving one equation for one variable (e.g., x or y) and substituting the resulting expression into the other equation. This reduces the system to a single equation with one variable, which can then be solved.

3x + 3y = 6

3(x+y) = 6

x+y = 6/3 = 2

Move y to right hand side to get x value

x = 2 - y

Now substitute x value in 2nd equation

6x - 4y = 18

6(2-y) - 4y = 18

12-6y-4y = 18

12-10y = 18

-10y = 18-12 = 6

y = - 6/10 = - 3/5

Now substitute y value in equation x = 2-y = 2- (-3/5 ) = 2 + 3/5 = 13/5

Method 2: Using Elimination Method

In this approach, you manipulate the equations (e.g., multiply, add, or subtract) to eliminate one variable. This simplifies the system to a single equation, which can then be solved for the remaining variable.

Compare x and y coefficients and make equal for one variable. Then, add or subtract the given equations to eliminate any one variable. Given equations are

3x + 3y = 6 and 6x - 4y = 18

Now multiply the 1st equation by 2 to make any one variable look equal.

2(3x + 3y) = 2(6)

6x + 6y = 12

Now subtract the 2nd equation from the above equation.

6x + 6y = 12

6x - 4y = 18

___________

10y = -6

y = -6/10 = -⅗

Now put y = -⅗ into the equation  3x + 3y = 6 

3x + 3 (-⅗) = 6

3x - 9/5 = 6

3x = 6 + 9/5 = 39/5

x = 13/5

Example 2:

Quantitative Comparison

2x + 4y = 6

x-y = 5

Quantity A: x

Quantity B: y

Possible Answers: 

A) Quantity A is greater B) The two quantities are equal C) Quantity B is greater

Solution Explanation:

Given equations are 2x + 4y = 6 and  x-y = 5

Let’s solve y using 2nd equation, 

x-y = 5

x = y+5

Substitute this in equation 1

2x + 4y = 6

2(y+5) + 4y = 6

2y + 10 + 4y = 6

6y + 10 = 6

Now combine the like terms

6y = 6-10 

6y = -4

y = -4/6 = -⅔

Now, substitute this value into equation x = y+5 to find the value of x.

x = y+5 = (-⅔) + 5 = (-2 + 15) / 3 = 13/3

The x value is positive and the y value is negative. Clearly, x is larger. Hence, the answer is A) Quantity A is greater.

GRE Linear Equations: Practice Questions & Answers

Practice Question 1:

Two cars start 20 miles apart and drive away from each other in opposite directions at speeds of 50 and 70 miles per hour.  In approximately how many minutes will they be 500 miles apart?

Possible Answers:

a) 200      b) 4    c) 240      d) none of these

Correct Answer: c

Answer Explanation:

Given, initial distance = 20,

Combined speed = 50 + 70 = 120

t is the number of hours.

The cars have  a distance from each other of 20 + 120t

Solve this equation for 500 miles

20 + 120t = 500

120t = (500-20) = 480

t = 480/120 = 4 hours

In the question it was asked in minutes, hence multiply the value of t by 60

t = 4 x 60 = 240 minutes

Practice Question 2:

x + y = 12 and 3x - y = 8

Quantity A: x

Quantity B: y

Possible Answers:

a) Quantity A is greater   b) Quantity B is greater   c) The two quantities are equal    d) the relationship can not be determined by the given information

Correct Answer: b

Answer Explanation:

Given equations are x + y = 12 and 3x - y = 8

x + y = 12

y = 12 - x

Now substitute this value in equation  3x - y = 8 and find the value of x

3x - y = 8

3x - (12-x) = 8

3x-12+x = 8

4x = 8+12 = 20

x = 20/4 = 5

Now substitute x=5 in the equation y = 12-x and determine the value of y

y = 12-x = 12-5 = 7

Quantity A: x=5  ; Quantity B: y =7

Quantity B is greater.

Practice Question 3:

Find the intersection of the following equations

2x + 4y = 6

16x - 4y = 4

Possible Answers:

a) (1,0.5)  b) (0.55,1.225)  c) (3,4)   d) (18,0)

Correct Answer: b

Answer Explanation:

Add both the equations and get the value of x

2x + 4y = 6

16x - 4y = 4

18x = 10

x = 10/18=5/9 = 0.55

Substitute x value in the equation 2x + 4y = 6

2(0.55) + 4y = 6

4y = 6-1.1 =4.9

y = 4.9/4=1.225

hence , (x,y) = (0.55,1.225)

Practice Question 4:

Sachi is 2 years younger than Ali

Dia is 5 years older than Tina

Ali is 6 years older than Tina

A---- Sachi Age

B----Dia's age

Possible Answers:

a) Quantity B is greater

b) The two quantities are equal

c) Quantity A is greater

d) The relationship cannot be determined

Correct Answer: B

Answer Explanation:

To simplify the word problem, you can express the given ages in terms of variables in a system of equations. Note that we have to compare S with D

S = A – 2

D = T + 5

A = T + 6

By substituting A for T in the first equation, we can get S in terms of T, which will let us directly compare the values of S and D.

S = (T + 6) – 2 = T + 4

If D = T + 5, and S = T + 4, D must be the greater value and Dia is one year older than Sachi.  B is the correct answer.

Practice Question 5:

If a = ⅓ b and b = 4c, then in terms of c, what is the value of a-b+c=?

Possible answers:

a) -11/3 c   b) -5/3 c     c) 5/3 c     d) c

Correct Answer: b

Answer Explanation:

Given, a = ⅓ b and b = 4c

Hence, a = ⅓ (4c) = 4/3 c 

Now substitute a=4/3 c and b= 4c in the given equation 

a-b+c = 4/3c -4c + c = 4c-12c+3c / 3 = -5/3 c

Practice Question 6:

If 3x+y = 14 and x-2y = -12, what is the value of x?

a) 1    b) ⅔     c) ⅓     d)   16/7

Correct Answer: d

Answer Explanation:

To solve this system of equations, we can use either elimination or substitution method. Now we are going to use the elimination method. 

3x+y = 14—----(1)

x-2y = -12—-----(2)

Now multiply eq(1) with 2, we get 2(3x+y) = 2(14)

6x+2y = 28—--(3)

Now add eq(3) and eq(2) and determine x value.

6x + 2y = 28

   x -2y  = -12

____________

7x = 16

  x = 16/7

Practice Question 7:

If 4(3x+y) = 16,   what is x in terms of y?

a) x = 3-3y    b) x = (10-y)/3   c) x = 1- y/3   d) y-4 / 3

Correct Answer: d

Answer Explanation:

Isolate x by using inverse operations. 

4(3x+y) = 16

3x+y = 16/4 = 4

Now move ‘y’ to right-hand side

3x+y = 4

3x = y-4

x = (y-4) / 3

Practice Question 8:

If 11+3x is  29, what is 4x?

a) 24   b) 12    c) 6     d) 2

Correct Answer: a

Answer Explanation:

First, we need to find the value of x

11+3x = 29

3x = 29-11 = 18

x = 18/3 = 6.

Now, 4x = 4*6 = 24.

Practice Question 9:

Raj sells bikes. His monthly salary is $1000. He gets a $500 commission for each bike he sells. If Raj wants to make $7,500 this month, how many bikes would he have to sell?

a) 14     b) 13     c) 10      d) 12

Correct Answer: b

Answer Explanation:

Let us assume, y = money earned and x = number of bikes sold

Here is the equation from the above data,  y = 500x + 1000

Substitute y = 7500 and calculate x

7500 = 500x + 1000

7500 - 1000  = 500x

6500 = 500x

6500/500 = x

13 = x

Hence, x = number of bikes sold = 13

Practice Question 10:

Joy bought some art supplies.  She bought colored pencils for $1.25 per box and sketch pads for $2.25 each.  Joy bought one more sketch pad than colored pencil boxes and spent $9.25.  How many sketch pads did she buy?

a) 2    b) 4     c) 3     d) 5

Correct Answer: c

Answer Explanation:

Let, x = no.of colour pencil boxes and x+1 = no.of sketch pads purchased

(Note: logically, 1.25+1 = 2.25 (matches with the sketch pads number given in question), so we used term x+1 for sketch pads)

The final equation becomes 1.25x + 2.25 (x+1) = 9.25. Now let’s, solve this equation

1.25x + 2.25 (x+1) = 9.25

1.25x + 2.25x + 2.25 = 9.25

Now combine the like terms on each side of the equation

3.5x = 9.25-2.25 =7

x= 7/3.5 = 2

Hence, the number of colour pencil boxes = 2

Number of sketch pads = x+1 = 2+1 = 3

Practice Question 11

In a group of 30 children, there are twice as many girls as there are boys. How many boys are there?

a) 14   b) 10    c) 8   d) 12

Correct Answer: b

Answer Explanation:

let , b= boys, g= girls

Given, that there are twice as many girls as there are boys, so 2b = g

Total = b+g = 30

Substitute g=2b in the above equation

b+2b=30

3b = 30

b = 30/3 = 10

Number of boys b = 10

Time-Saving Techniques for Solving GRE Linear Equations

Here are time-saving techniques for solving linear equations in the GRE:

• Combine like terms immediately: Simplify the equation by merging similar terms on both sides.
• Eliminate fractions: Multiply through by the least common denominator (LCD) to remove fractions.
• Keep variables on one side: Move all variables to one side and constants to the other.
• Substitute in word problems: Assign variables to unknowns and solve using substitution for clarity.
• Skip unnecessary steps: Combine terms and simplify directly to avoid redundant writing.
• Backsolve for multiple-choice: Test answer choices (start with the middle value) to find the correct solution.
• Practice mental math: Develop quick calculation skills to save time on basic operations.
• Use estimation: Verify solutions by approximating when possible.
• Write clearly and concisely: Avoid mistakes caused by messy or unclear steps.