Achievable ASVAB

2. Math Knowledge

2.2. Algebra II

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Systems of equations

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Definitions

System of equations: A set of multiple equations with multiple variables

A system of equations is a set of two or more equations with multiple variables you need to identify. The most common type of system problem on the ASVAB is a two-variable system with two equations.

For example, let’s say we have the system:

$2 x + 5 y = - 1$

$- 3 x - 10 y = 4$

We want to solve for the values of both $x$ and $y$ .

The answer to this system is $x = 2$ and $y = - 1$ , but how did we get there?

There are multiple ways to solve a systems problem, but let’s look at the main two: elimination and substitution.

Elimination method

Definitions

Elimination: Cancelling out a variable in a system of equations

The elimination method is a way to solve a system by getting rid of one variable so you can solve for the other. We do this by lining up the equations and adding or subtracting them until one variable cancels out.

Signs in elimination: The sign in front of the value doesn’t matter. If they have the same sign, subtract the second equation from the first. If they have opposite signs, add the two equations.

Example:

To use elimination, we line up the equations vertically and then either add or subtract them to eliminate one of the variables. So let’s first line them up.

$2 x + 5 y = - 1$

$- 3 x - 10 y = 4$

Our goal is to get the coefficients for either $x$ or $y$ to match so we can eliminate that variable. For example, we can eliminate $y$ by getting both equations to have a $+ / - 10 y$ term.

In this case, we can get the top equation with $5 y$ to become $10 y$ by multiplying by $2$ . Be cautious! Whenever we multiply one value in an equation, we must multiply each value in that equation by the same amount.

$2 (2 x + 5 y = - 1) \to (4 x + 10 y = - 2)$

Now line them up:

$4 x + 10 y = - 2$

$- 3 x - 10 y = 4$

The $y$ terms now have the same value with opposite signs. Because their signs are opposite, we’ll add both equations to eliminate $y$ .

$4 x + 10 y = - 2$

$- 3 x - 10 y = 4$

$4 x - 3 x = - 2 + 4$

$x = 2$

Now we know the value of one variable.

To find $y$ , plug $x = 2$ into one of the original equations:

$2 x + 5 y = - 1$

$2 (2) + 5 y = - 1$

$4 + 5 y = - 1$

$5 y = - 5$

$y = - 1$

We’ve found both values: $x = 2$ and $y = - 1$ .

Knowledge check:

Solve using elimination:

$x + y x - y = 6 = 2$

(spoiler)

$x = 4, y = 2$

Substitution method

Definitions

Substitution: Isolation and replacing a variable in a system of equations

As an alternative to elimination, we can use substitution.

To use substitution, we solve one equation for one variable and then plug that expression into the other equation.

Example:

$2 x + 4 y = 14$

$5 x - 2 y = - 25$

First, isolate $x$ in the first equation:

$2 x + 4 y = 14$

$2 x = - 4 y + 14$

$x = - 2 y + 7$

Now replace $x$ in the second equation:

$5 x - 2 y = - 25$

$5 (- 2 y + 7) - 2 y = - 25$

Now we have one variable:

$- 10 y + 35 - 2 y = - 25$

$- 12 y = - 60$

$y = 5$

Now plug $y = 5$ back in:

$x = - 2 y + 7$

$x = - 2 (5) + 7$

$x = - 10 + 7$

$x = - 3$

So the solution is $x = - 3$ and $y = 5$ .

Knowledge check:

Solve using substitution:

$x 2 x + y = y + 1 = 7$

(spoiler)

$x = \frac{8}{3}, y = \frac{5}{3}$

When to use each method

Some people prefer one method over the other, while others choose based on the problem.

Here are some quick tips:

If one variable is easy to isolate, use substitution
If coefficients already match (or are easy to match), use elimination
If only one equation needs multiplying, elimination is usually easier
If both equations need multiplying, substitution may be simpler

Knowledge check:

Which method would be easiest to use for this problem?

$x 3 x + y = 2 y - 3 = 10$

(spoiler)

Substitution, because one variable is already isolated.

Next | 2.2.2 Functions and complex numbers

Systems of equations

Definitions

System of equations: A set of multiple equations with multiple variables

For example, let’s say we have the system:

$2 x + 5 y = - 1$

$- 3 x - 10 y = 4$

We want to solve for the values of both $x$ and $y$ .

The answer to this system is $x = 2$ and $y = - 1$ , but how did we get there?

There are multiple ways to solve a systems problem, but let’s look at the main two: elimination and substitution.

Elimination method

Definitions

Elimination: Cancelling out a variable in a system of equations

Signs in elimination: The sign in front of the value doesn’t matter. If they have the same sign, subtract the second equation from the first. If they have opposite signs, add the two equations.

Example:

To use elimination, we line up the equations vertically and then either add or subtract them to eliminate one of the variables. So let’s first line them up.

$2 x + 5 y = - 1$

$- 3 x - 10 y = 4$

Our goal is to get the coefficients for either $x$ or $y$ to match so we can eliminate that variable. For example, we can eliminate $y$ by getting both equations to have a $+ / - 10 y$ term.

In this case, we can get the top equation with $5 y$ to become $10 y$ by multiplying by $2$ . Be cautious! Whenever we multiply one value in an equation, we must multiply each value in that equation by the same amount.

$2 (2 x + 5 y = - 1) \to (4 x + 10 y = - 2)$

Now line them up:

$4 x + 10 y = - 2$

$- 3 x - 10 y = 4$

The $y$ terms now have the same value with opposite signs. Because their signs are opposite, we’ll add both equations to eliminate $y$ .

$4 x + 10 y = - 2$

$- 3 x - 10 y = 4$

$4 x - 3 x = - 2 + 4$

$x = 2$

Now we know the value of one variable.

To find $y$ , plug $x = 2$ into one of the original equations:

$2 x + 5 y = - 1$

$2 (2) + 5 y = - 1$

$4 + 5 y = - 1$

$5 y = - 5$

$y = - 1$

We’ve found both values: $x = 2$ and $y = - 1$ .

Knowledge check:

Solve using elimination:

$x + y x - y = 6 = 2$

(spoiler)

$x = 4, y = 2$

Substitution method

Definitions

Substitution: Isolation and replacing a variable in a system of equations

As an alternative to elimination, we can use substitution.

To use substitution, we solve one equation for one variable and then plug that expression into the other equation.

Example:

$2 x + 4 y = 14$

$5 x - 2 y = - 25$

First, isolate $x$ in the first equation:

$2 x + 4 y = 14$

$2 x = - 4 y + 14$

$x = - 2 y + 7$

Now replace $x$ in the second equation:

$5 x - 2 y = - 25$

$5 (- 2 y + 7) - 2 y = - 25$

Now we have one variable:

$- 10 y + 35 - 2 y = - 25$

$- 12 y = - 60$

$y = 5$

Now plug $y = 5$ back in:

$x = - 2 y + 7$

$x = - 2 (5) + 7$

$x = - 10 + 7$

$x = - 3$

So the solution is $x = - 3$ and $y = 5$ .

Knowledge check:

Solve using substitution:

$x 2 x + y = y + 1 = 7$

(spoiler)

$x = \frac{8}{3}, y = \frac{5}{3}$

When to use each method

Some people prefer one method over the other, while others choose based on the problem.

Here are some quick tips:

If one variable is easy to isolate, use substitution
If coefficients already match (or are easy to match), use elimination
If only one equation needs multiplying, elimination is usually easier
If both equations need multiplying, substitution may be simpler

Knowledge check:

Which method would be easiest to use for this problem?

$x 3 x + y = 2 y - 3 = 10$

(spoiler)

Substitution, because one variable is already isolated.

Definitions

System of equations: A set of multiple equations with multiple variables

For example, let’s say we have the system:

$2 x + 5 y = - 1$

$- 3 x - 10 y = 4$

We want to solve for the values of both $x$ and $y$ .

The answer to this system is $x = 2$ and $y = - 1$ , but how did we get there?

There are multiple ways to solve a systems problem, but let’s look at the main two: elimination and substitution.

Elimination method

Definitions

Elimination: Cancelling out a variable in a system of equations

Signs in elimination: The sign in front of the value doesn’t matter. If they have the same sign, subtract the second equation from the first. If they have opposite signs, add the two equations.

Example:

To use elimination, we line up the equations vertically and then either add or subtract them to eliminate one of the variables. So let’s first line them up.

$2 x + 5 y = - 1$

$- 3 x - 10 y = 4$

Our goal is to get the coefficients for either $x$ or $y$ to match so we can eliminate that variable. For example, we can eliminate $y$ by getting both equations to have a $+ / - 10 y$ term.

In this case, we can get the top equation with $5 y$ to become $10 y$ by multiplying by $2$ . Be cautious! Whenever we multiply one value in an equation, we must multiply each value in that equation by the same amount.

$2 (2 x + 5 y = - 1) \to (4 x + 10 y = - 2)$

Now line them up:

$4 x + 10 y = - 2$

$- 3 x - 10 y = 4$

The $y$ terms now have the same value with opposite signs. Because their signs are opposite, we’ll add both equations to eliminate $y$ .

$4 x + 10 y = - 2$

$- 3 x - 10 y = 4$

$4 x - 3 x = - 2 + 4$

$x = 2$

Now we know the value of one variable.

To find $y$ , plug $x = 2$ into one of the original equations:

$2 x + 5 y = - 1$

$2 (2) + 5 y = - 1$

$4 + 5 y = - 1$

$5 y = - 5$

$y = - 1$

We’ve found both values: $x = 2$ and $y = - 1$ .

Knowledge check:

Solve using elimination:

$x + y x - y = 6 = 2$

(spoiler)

$x = 4, y = 2$

Substitution method

Definitions

Substitution: Isolation and replacing a variable in a system of equations

As an alternative to elimination, we can use substitution.

To use substitution, we solve one equation for one variable and then plug that expression into the other equation.

Example:

$2 x + 4 y = 14$

$5 x - 2 y = - 25$

First, isolate $x$ in the first equation:

$2 x + 4 y = 14$

$2 x = - 4 y + 14$

$x = - 2 y + 7$

Now replace $x$ in the second equation:

$5 x - 2 y = - 25$

$5 (- 2 y + 7) - 2 y = - 25$

Now we have one variable:

$- 10 y + 35 - 2 y = - 25$

$- 12 y = - 60$

$y = 5$

Now plug $y = 5$ back in:

$x = - 2 y + 7$

$x = - 2 (5) + 7$

$x = - 10 + 7$

$x = - 3$

So the solution is $x = - 3$ and $y = 5$ .

Knowledge check:

Solve using substitution:

$x 2 x + y = y + 1 = 7$

(spoiler)

$x = \frac{8}{3}, y = \frac{5}{3}$

When to use each method

Some people prefer one method over the other, while others choose based on the problem.

Here are some quick tips:

If one variable is easy to isolate, use substitution
If coefficients already match (or are easy to match), use elimination
If only one equation needs multiplying, elimination is usually easier
If both equations need multiplying, substitution may be simpler

Knowledge check:

Which method would be easiest to use for this problem?

$x 3 x + y = 2 y - 3 = 10$

(spoiler)

Substitution, because one variable is already isolated.