# Simultaneous Equations - GMAT Math Study Guide

## Definitions

• Simultaneous Equations - Also known as a system of equations, simultaneous equations are a set of equations containing multiple variables.
For example, the equations x + 2y = 10 and 3x - y = -3 form simultaneous equations.
• Solution - A value or set of values that make the simultaneous equations true.
For example, the solution to the equations 2x + 3y = 15 and x - y = -10 is x = -3 and y = 7.

Simultaneous equations are used when trying to find the intersection of two lines (two equations) or three planes (three equations). If any of the equations are equivalent, there will be an infinite number of solutions. If none of the equations are equivalent, then either one unique solution or no solution exists. (The process of solving simultaneous equations and finding whether a solution exists will be discussed in much greater detail in a few paragraphs).

### Infinite Solutions

Two equations with two unknowns do not always have a unique solution.

You may remember the axiom: if you have 2 equations and 2 unknowns, you can find a solution. While this is technically true, it can be easily manipulated to trip up students since equivalent equations are actually the same line. Thus, there are an infinite number of solution points. Since this can be confusing in the abstract, consider the following example:

Equation 1: x + 2y = 5
Equation 2: -10 + 4y = -2x

Equation 1, Multiplied by 2: 2x + 4y = 10
Equation 1, Multiplied by 2: 2x + 4y - 10 = 10 - 10
Equation 1, Multiplied by 2: 2x + 4y - 10 = 0
Equation 1, Multiplied by 2: 2x - 2x + 4y - 10 = 0 - 2x
Equation 1, Multiplied by 2: 4y - 10 = - 2x
Equation 1, Multiplied by 2: - 10 + 4y = - 2x
Notice that Equation 1 = Equation 2

### One Solution

Equation 1: x - y = -10
Equation 2: 2x + 3y = 15

Equation 1, Rearranged: y = x + 10
Equation 1 Substituted into 2: 2x + 3(x + 10) = 15
Equation 1 Substituted into 2: 2x + 3x + 30 = 15
Equation 1 Substituted into 2: 5x + 30 = 15
Equation 1 Substituted into 2: 5x + 30 - 30 = 15 - 30
Equation 1 Substituted into 2: 5x = -15
Equation 1 Substituted into 2: x = -3
Equation 1, Solve for Y: 2(-3) + 3(y) = 15
Equation 1, Solve for Y: -6 + 3y = 15
Equation 1, Solve for Y: 3y = 21
Equation 1, Solve for Y: y = 7
Solution: x = -3, y = 7

### No Solution

Equation 1: 6x + 2y = 10
Equation 2: 12x + 4y = 21

Equation 1, Multiplied by 2: 12x + 4y = 20

The first equation contradicts the second equation. Thus, there will be no solution for these equations as no x and y will satisfy both equations. This happens because the two equations are parallel lines but have different x-intercepts and thus never intersect.

## Solving Simultaneous Equations

### What is a Solution?

For some students, the concept of a solution seems abstract and amorphous. Mathematically, a solution is simply a pair of points that make a series of equations true. Graphically represented, a solution is a series of points where the lines (or planes, when dealing with three equations) intersect.

The graph below depicts y = 2x [the red line] and y = x + 2 [the blue line]. These two equations form a series of simultaneous equations. The solution to this set of simultaneous equations is x = 2 and y = 4. From a mathematical point of view, these numbers are solutions because the value for x and the value for y make both equations simultaneously true. In other words, x = 2 and y = 4 is the only combination of values that satisfy both equations simultaneously. Graphically, the solution is represented by the intersection of the lines.

Equation 1 [Red Line]: y = 2x
Equation 2 [Blue Line]: y = x + 2

Solution: x = 2, y = 4

### Solving by Substitution

Use Case: When one variable can be easily isolated in one of the two equations.

1. Solve the first equation for one variable.
2. Substitute this solved equation into the second equation. Solve the second equation for the second variable. This solution for the second variable should be a constant (not a variable).
3. Substitute the solution for the second variable back into the first equation and solve for the first variable.
4. Test the solution by substituting it into both of the equations and making sure the equations hold true.

These steps may not make sense in the abstract, so we will see how they work with examples.

Equation 1: 4y = 12x + 20
Equation 2: 4x + 2y = 20

Since one variable (i.e., y in equation 1) can be easily isolated, solve by substitution.

Step 1: Equation 1, Solved for Y:
y = 3x + 5 [Divided by 4]

Step 2:
Equation 2, Substituted Y1: 4x + 2(3x + 5) = 20
Equation 2, Substituted Y1: 4x + 6x + 10 = 20
Equation 2, Substituted Y1: 10x + 10 = 20
Equation 2, Substituted Y1: 10x + 10 - 10 = 20 - 10
Equation 2, Substituted Y1: 10x = 10
Equation 2, Substituted Y1: x = 1

Step 3:
4y = 12(1) + 20
4y = 32
y = 8

Step 4: Test the Solutions [x = 1, y = 8]
Equation 1: 4(8) =? 12(1) + 20
32 =? 20 + 12
32 = 32
Equation 2: 4(1) + 2(8) =? 20
4(1) + 2(8) =? 20
4 + 16 =? 20
20 = 20

Another Example:

Equation 1: 2x + 6y = 18
Equation 2: 5x = 10y + 20

Since one variable (i.e., x in equation 2) can be easily isolated, solve by substitution.

Step 1: Equation 2, Solved for X:
x = 2y + 4 [Divided by 5]

Step 2:
Equation 1, Substituted X2: 2(2y + 4) + 6y = 18
Equation 1, Substituted X2: 4y + 8 + 6y = 18
Equation 1, Substituted X2: 4y + 8 - 8 + 6y = 18 - 8
Equation 1, Substituted X2: 4y + 6y = 10
Equation 1, Substituted X2: 10y = 10
Equation 1, Substituted X2: y = 1

Step 3:
5x = 10(1) + 20
5x = 10 + 20
5x = 30
x = 6

Step 4: Test the Solutions [x = 6, y = 1]
Equation 1: 2(6) + 6(1) =? 18
12 + 6 =? 18
18 = 18

Equation 2: 5(6) =? 10(1) + 20
30 =? 10 + 20
30 = 30

### Solving by Subtraction or Addition

Use Case: There is an easy way to manipulate both equations by multiplication and end up with one of the variables having the same coefficient in both equations.

1. Multiply the two equations so that one variable has the same coefficient in both equations (ignoring sign).
2. Add or subtract the two equations so that the variable with the same coefficient will cancel out.
3. Solve for the variable that remains.
4. Substitute the solution for the variable into one of the original equations and solve for the remaining variable.
5. Test the solution by substituting it into both of the equations and making sure the equations hold true.

These steps probably do not make sense in abstract, so we will see how they work with examples.

Equation 1: 2x + 3y = 10
Equation 2: 4x + y = 5

Since one equation (i.e., Equation 1) can be easily manipulated so that the coefficients of a variable are equal, solve by subtraction.

Step 1: Multiply to Reach Equal Coefficients
Equation 1 (times 2): 4x + 6y = 20

Both equations have an x coefficient of 4.

Step 2: Subtract and Eliminate a Variable
4x + 6y = 20
(4x + y  = 5)
4x - 4x + 6y - y = 20 - 5
5y = 15

Step 3: Solve for the Remaining Variable
5y = 15
y = 3

Step 4: Substitute the Equation
2x + 3(3) = 10
x = 1/2

Step 5: Test the Solutions [x = 1/2, y = 3]
Equation 2: 4(1/2) + (3) =? 5
2 + (3) =? 5
5 = 5
Equation 1: 2(1/2) + 3(3) =? 10
1 + 9 =? 10
10 = 10

Equation 1: 5x + 2y = 8
Equation 2: -3x - y = 1

Since one equation (i.e., Equation 2) can be easily manipulated so that the coefficients of a variable are equal, solve by addition.
Step 1: Multiply to Reach Equal Coefficients
Equation 2 (times 2): -6x - 2y = 2

Step 2: Add and Eliminate a Variable
5x + 2y = 8
+ (-6x - 2y = 2)
5x + (-6x) + 2y + (-2y) = 8 + 2

Step 3: Solve for the Remaining Variable
-x = 10
x = -10

Step 4: Substitute the Equation
-3(-10) - y = 1
30 - y = 1
-y = -29
y = 29

Step 5: Test the Solutions [x = -10, y = 29]
Equation 1: 5(-10) + 2(29) =? 8
-50 + 58 =? 8
8 = 8
Equation 2: -3(-10) - (29) =? 1
30 - 29 =? 1
1 = 1

## Trap: Duplicate Equations

While the common trap of placing two equations that look different but are actually similar received attention earlier in this study guide, this trap snags so many students that it deserves a little more coverage.

The crux of the trap is that two equations are made to look different when in reality they are the same. A single line (e.g., y = 2x + 3) will have an infinite number of solutions (i.e., pairs of x and y coordinates that satisfy it) and as a result, there is no solution.

Equation 1: y = 2x + 3
Equation 2: -15 = -5(y - 2x)
Equation 2, Manipulated: -15 = -5y + 10x
Equation 2, Manipulated: 5y = 15 + 10x
Equation 2, Manipulated: y = 3 + 2x
Equation 2, Manipulated: y = 2x + 3

Since the equations are the same, there is no definitive x,y pair that is a solution. Instead, there are an infinite number of solutions.

## Types of GMAT Problems

1. Solving A Set of Two Equations Using Substitution

Solving by substitution should be used when one variable can be easily isolated in one of the two equations. Recall that solving for two equations is solving for their intersection. Thus, the x and y values must make both equations true. Begin by solving the first equation for one variable. This "solution" will involve both constant terms as well as the other variable. Substitute this solution into the second equation. Solve the second equation for the second variable. This solution for the second variable should be a constant. Substitute the solution for the second variable back into the first equation and solve for the first variable. To test the solutions, substitute them both into one of the equations and make sure the statement is true.

Solve the following system of equations using substitution:
9x+6y = 30
6x-4y = 4
 A) x=5, y=6 B) x=2, y=3 C) x=2, y=2 D) x=1, y=2 E) x=3, y=3
1. Solve the first eqatuion for x.
9x+6y = 30
9x = 30-6y
x = (30-6y)/9
2. Substitute the solution for x into the second equation and solve for y.
6x-4y = 4
6*((30-6y)/9)-4y = 4
((180-36y)/9)-4y = 4
(20-4y) -4y = 4
-8y = -16
y=2
3. Substitute the solution for y into the first equation and solve for x.
9x+6y = 30
9x+6*2 = 30
9x+12 = 30
9x = 18
x=2
4. Check the answer by substituting x=2, y=2 into the first equation.
9*2+6*2 = 30
18+12 = 30
30 = 30
2. Solving A Set of Two Equations Using Subtraction

Solving by subtraction should be used when there is an easy way to manipulate both equations by multiplication and end up with one of the variables having the same coefficient in both equations. Begin by multiplying the two equations so that one variable has the same coefficient in both equations (ignoring sign). Add or subtract the two equations so that the variable with the same coefficient will cancel out. Solve for the variable that remains. Finally, substitute the solution for the variable into one of the original equations and solve for the remaining variable.

Solve the set of linear equations using subtraction:
4x-3y = 3
2x+6y = 4
 A) x=1/3, y=2/3 B) x=1, y=2/3 C) x=3, y=1/3 D) x=2, y=2 E) x=1, y=1/3
1. Multiply the first equation by 2 so y will have a coefficient of 6 in both equations.
4x-3y = 3
Multiplying by 2: 8x-6y = 6
2. Add the new equation with the original second equation and solve for x.
8x-6y = 6
+2x+6y = 4
10x = 10
x=1
3. Substitute the solution for x into the first equation and solve for y.
4x-3y = 3
4(1)-3y = 3
-3y = -1
y=1/3
4. Substitute the solutions into the second equation to check the answer.
2*1+6*(1/3) ?= 4
2+(6/3) ?= 4
2+2 ?= 4
4=4
5. x = 1 and y=1/3
3. Solving for Three Simultaneous Equations

The process is very similar to solving for two equations. Either substitution or subtraction can be used, or a mix of both. However, after solving for one variable, it must be substituted into both of the other equations. Once this initial substitution is complete, the problem has been reduced to a system of two equations. Once the solutions for both of the variables in this new system of two equations are found, the final variable may be found by substituting the two solutions into any of the three original equations.

Solve the following system of equations:
x-y-z = 0
3x+4y+3z = 4
5x+2y+7z = 22
 A) x=1, y=2, z=3 B) x=-1, y=-2, z=1 C) x=-2, y=3, z=1 D) x=1, y=-2, z=3 E) x=3, y=-1, z=2
1. Solve the first equation for x.
x-y-z = 0
x = y+z
2. Substitute the solution for x into the second and third equations.
3*(y+z)+4y+3z = 4
3y+3z+4y+3z = 4
7y+6z = 4

5*(y+z)+2y+7z = 22
5y+5z+2y+7z = 22
7y+12z = 22
3. Subtract the new second equation from the new third equation and solve for z.
7y+12z = 22
- (7y+6z = 4)
6z = 18
z=3
4. Substitute the solution for z back into 7y+6z = 4 and solve for y.
7y+6*3 = 4
7y+18 = 4
7y = -14
y=-2
5. Substitute the values for y and z back into x-y-z = 0.
x-(-2)-3 = 0
x-1 = 0
x=1